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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

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CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Students can easily download these questions in PDF format and refer to them for exam preparation.

CBSE Class 9 All Students can also Download here Class 9 Other Study Materials in PDF Format.

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CBSE Class 9 Mathematics Case Study Questions

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If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

  • Now he told Raju to draw another line CD as in the figure
  • The teacher told Ajay to mark  ∠ AOD  as 2z
  • Suraj was told to mark  ∠ AOC as 4y
  • Clive Made and angle  ∠ COE = 60°
  • Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

  • 2y + z = 90°
  • 2y + z = 180°
  • 4y + 2z = 120°
  • (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

  • (a) 31.6 m²
  • (c) 513.3 m³
  • (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

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Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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12 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

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Case Study Questions for Class 9 Maths

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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.

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CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

Table of Contents

CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution

Case study questions are a form of examination where students are presented with real-life scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problem-solving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.

Chapterwise Case Study Questions for Class 9 Maths

Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to real-world situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problem-solving strategies.

  • Case Study Questions for Chapter 1 Number System
  • Case Study Questions for Chapter 2 Polynomials
  • Case Study Questions for Chapter 3 Coordinate Geometry
  • Case Study Questions for Chapter 4 Linear Equations in Two Variables
  • Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
  • Case Study Questions for Chapter 6 Lines and Angles
  • Case Study Questions for Chapter 7 Triangles
  • Case Study Questions for Chapter 8 Quadilaterals
  • Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
  • Case Study Questions for Chapter 10 Circles
  • Case Study Questions for Chapter 11 Constructions
  • Case Study Questions for Chapter 12 Heron’s Formula
  • Case Study Questions for Chapter 13 Surface Area and Volumes
  • Case Study Questions for Chapter 14 Statistics
  • Case Study Questions for Chapter 15 Probability

The above  Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.

  • Class 9 Science Case Study Questions
  • Class 9 Social Science Case Study Questions

How to Approach Case Study Questions

When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:

  • Read the case study carefully: Understand the given scenario and identify the key information.
  • Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
  • Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
  • Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
  • Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.

Tips for Solving Case Study Questions

Here are some valuable tips to help you effectively solve case study questions:

  • Read the question thoroughly and underline or highlight important information.
  • Break down the problem into smaller, manageable parts.
  • Visualize the problem using diagrams or charts if applicable.
  • Use appropriate mathematical formulas and concepts to solve the problem.
  • Show all the steps of your calculations to ensure clarity.
  • Check your final answer and review the solution for accuracy and relevance to the case study.

Benefits of Practicing Case Study Questions

Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:

  • Enhances critical thinking skills
  • Improves problem-solving abilities
  • Deepens understanding of mathematical concepts
  • Develops analytical reasoning
  • Prepares you for real-life applications of mathematics
  • Boosts confidence in approaching complex mathematical problems

Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problem-solving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.

Q1: Can case study questions help me score better in my Class 9 Maths exams?

Yes, practicing case study questions can significantly improve your problem-solving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their real-life applications.

Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?

Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.

Q3: Are the solutions provided for the case study questions in the PDF resource?

Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problem-solving process.

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Chapter 2 Class 9 Polynomials

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Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Answers to each and every question is explained in an easy to understand way, with videos of all the questions.

In this chapter, we will learn

  • What is a Polynomial
  • What are Polynomials in One Variable
  • What are monomials, binomials, trinomials
  • What is the Degree of a Polynomial
  • What are Linear, Quadratic and Cubic Polynomials
  • What is zero of a polynomial (or root of a polynomial)
  • Finding zeroes of a polynomial
  • Dividing Polynomials and finding remainder
  • Finding Remainder using Remainder Theorem
  • Checking if it is a factor or not
  • Factorising Quadratic and Cubic Polynomials using Factor Theorem
  • Solving questions using Algebra Identities (Check full list of Algebra Formulas )

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  • Math Article
  • Polynomials Class 9

Polynomial Class 9 Notes - Chapter 2

Cbse class 9 maths polynomials notes:- download pdf here.

Polynomial is derived from the words ‘poly’, which means ‘many’, and the word ‘nomial’, which means ‘term’. In maths, a polynomial expression consists of variables which are also known as indeterminates and coefficients. The coefficients involve the operations of subtraction, addition, non-negative integer exponents of variables and multiplication. In mathematics, both algebraic expressions and polynomials are made up of variables and constants, including arithmetic operations. The only difference between them is that algebraic expressions contain irrational numbers in the powers. A detailed polynomials Class 9 notes are provided here along with some important questions so that students can understand the concept easily.

Polynomials Class 9 Notes

To prepare for Class 9 exams, students will require notes to study. These notes are of great help when they have to revise chapter 2 polynomials before the exam. The notes provide a brief of the chapter so that students find it easy to have a glance at once. Go through the key points given and solve problems based on them. The topics and subtopics covered in Class 9 polynomials chapter 2 include:

  • Introduction

Polynomials in One Variable

  • Zeros of Polynomials

Remainder Theorem

Factorisation of polynomials, algebraic identities, polynomial definition.

Polynomials are expressions with one or more terms with a non-zero coefficient. A polynomial can have more than one term. An algebraic expression p(x) = a 0 x n  + a 1 x n-1  + a 2 x n-2  + … a n is a polynomial where a 0 , a 1 , ………. a n are real numbers and n is non-negative integer. Examples of polynomials are:

  • w + x + y + z
  • x 2  + x + 1

To know more about polynomials, click here .

In the polynomial, each expression is called a term .   Suppose x 2 + 5x + 2 is polynomial, then the expressions x 2 , 5x, and 2 are the terms of the polynomial.  

Coefficient

Each term of the polynomial has a coefficient . For example, if 2x + 1 is the polynomial, then the coefficient of x is 2.

Types of Polynomial

A polynomial of 1 term is called a monomial. Example: 2x. A polynomial of 2 terms is called a binomial. Example: 5x + 2. A polynomial of 3 terms is called a trinomial. Example: 2x + 5y – 4.

Constant Polynomial

Real numbers can also be expressed as polynomials. 3, 6, and 7 are also polynomials without any variables. These are called constant polynomials .  The constant polynomial 0 is called zero polynomial . The exponent of the polynomial should be a whole number. For example,  x -2 + 5x + 2, cannot be considered a polynomial since the exponent of x is -2, which is not a whole number.

Degree of a Polynomial

The highest power of the polynomial is called the  degree of the polynomial . For example, in x 3  + y 3  + 3xy(x + y), the degree of the polynomial is 3. For a non-zero constant polynomial, the degree is zero. Apart from these, there are other types of polynomials such as:

  • Linear polynomial – of degree one
  • Quadratic Polynomial- of degree two
  • Cubic Polynomial – of degree three

This topic has been widely discussed in class 9 and class 10.

Polynomials in one variable are expressions which consist of only one type of variable in the entire expression. Example of polynomials in one variable:

  • 2x 2 + 5x + 15

To know more about polynomials in one variable, click here .

Zeroes of Polynomial

The zeroes of polynomials are the points where the polynomial is equal to 0 as a whole.

To know more about the zeroes of polynomials, click here .

If p(x) is any polynomial having degree greater than or equal to 1, and if it is divided by the linear polynomial x – a, then the remainder is p(a).

To know more about the Remainder Theorem, click here .

Factor Theorem

x – c is a factor of the polynomial p(x), if p(c) = 0. Also, if x – c is a factor of p(x), then p(c) = 0.

To know more about the Factor Theorem, click here .

Factorisation of polynomials is the process of expressing the polynomials as the product of two or more polynomials. For example, the polynomial x 2 -x-6 can be factorised as (x-3)(x+2)

Also read: Factorisation of Polynomials

Algebraic identities are algebraic equations which are valid for all values. The important algebraic identities used in Class 9 Maths chapter 2 polynomials are listed below:

  • (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx
  • (x + y) 3  = x 3  + y 3  + 3xy(x + y)
  • (x – y) 3 = x 3 – y 3 – 3xy(x – y)
  • x 3 + y 3 + z 3 – 3xyz = (x + y + z) (x 2 + y 2 + z 2 – xy – yz – zx)

To learn more about Algebraic Identities for class 9, Click here .

Explore more:

For more information on quadratic polynomial, watch the below video.

class 9 maths case study questions with solutions polynomials

Polynomials Class 9 Examples

Write the coefficients of x in each of the following:

  • 23x 2 – 5x + 1

In 3x + 1, the coefficient of x is 3.

In 23x 2 – 5x + 1, the coefficient of x is -5.

What are the degrees of following polynomials?

  • 3a 2 + a – 1
  • 32x 3 + x – 1
  • 3a 2 + a – 1 : The degree is 2
  • 32x 3 + x – 1 : The degree is 3

Polynomials Class 9 Important Questions

  • Find value of polynomial 2x 2 + 5x + 1 at x = 3.
  • Check whether x = -1/6 is zero of the polynomial p(a) = 6a + 1.
  • Divide 3a 2 + x – 1 by a + 1.
  • Find value of k, if (a – 1) is the factor of p(a) = ka 2 – 3a + k.
  • 4x 2 + 9y 2 + 16z 2 + 12xy – 24yx – 16xz
  • 2x 2 + y 2 + 8z 2 – 2√2xy + 4√2yz – 8xz

Stay tuned with BYJU’S – The Learning App and get detailed notes of all concepts of Class 9 Mathematics.

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  • NCERT Exemplar for Class 9 Maths Chapter 2 - Polynomials (Book Solutions)
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NCERT Exemplar for Class 9 Maths - Polynomials - Free PDF Download

Free PDF download of NCERT Exemplar for Class 9 Maths Chapter 2 - Polynomials solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2 - Polynomials exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations. Download free Class 9 Maths to amp up your preparations and to score well in your examinations. Students can also avail of NCERT Solutions Class 9 Science from our website. Besides, find NCERT Solutions to get more understanding of various subjects. The solutions are up-to-date and are sure to help in your academic journey.

Access NCERT Exemplar Solutions for Class 9 Mathematics Chapter 2 - Polynomials

Multiple choice questions.

Sample Question 1: If \[{x^2}{\text{ }} + {\text{ }}kx{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}3} \right)\] for all x, then the value of k is 

(A)  $ 1 $ 

(B)  $  - 1 $ 

(C)  $ 5 $ 

(D)  $ 3 $ 

Ans:  Option (C) is correct.

We have, \[{x^2}{\text{ }} + {\text{ }}kx{\text{ }} + {\text{ }}6{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}3} \right)\]

Now, let’s taking RHS,

 $  \Rightarrow \left( {x + 2} \right)\left( {x + 3} \right) $ 

On multiplication, we get

 $  \Rightarrow {x^2} + 2x + 3x + 6 $ 

On addition, we get

 $  \Rightarrow {x^2} + 5x + 6 $ 

 Now on comparing LHS and RHS, we get

 $ k = 5 $ 

EXERCISE 2.1

Write the correct option in each of the following:

 1. Which of the following is a polynomial?

(A)  $ \dfrac{{{x^2}}}{2} - \dfrac{2}{{{x^2}}} $ 

(B) $ \sqrt {2x}  - 1 $ 

(C)  $ {x^2} + \dfrac{{3{x^{\dfrac{3}{2}}}}}{{\sqrt x }} $ 

(D)  $ \dfrac{{x - 1}}{{x + 1}} $ 

Ans: (C) is the correct option.

Polynomials are an algebraic expression that consist of variables and coefficients. Here the power of variables should be in whole number like 1,2,3 etc.

 (A)  $ \dfrac{{{x^2}}}{2} - \dfrac{2}{{{x^2}}} $ 

Here the coefficient  $  - 2 $  is multiplied with  $ {x^{ - 2}} $ . So here the power of x is not a whole number. Hence, it is not a Polynomial.

Here the coefficient  $ \sqrt 2  $  is multiplied with  $ {x^{\dfrac{1}{2}}} $ . So here the power of x is not a whole number. Hence, it is not a Polynomial.

Here the coefficient  $ 1 $  is multiplied with  $ {x^2} $  and the coefficient  $ 3 $  is multiplied with \[{x^{\dfrac{3}{2} - \dfrac{1}{2}}} = {x^{\dfrac{2}{2}}} = x\]  . So here the power of x is a whole number in the whole equation. Hence, it is a Polynomial.

It is not a standard format of a Polynomial. Hence, it is not a Polynomial.

2. $ \sqrt 2  $ is a Polynomial of degree

(A)  $ 2 $ 

(B)  $ 0 $ 

(C)  $ 1 $ 

(D)  $ \dfrac{1}{2} $ 

Ans: Option (B) is correct.

Here  $ \sqrt 2  $  is a constant Polynomial. We can also write it as  $ \sqrt 2 {x^0} $ . Therefore, the degree of this Polynomial is  $ 0 $ .

3. Degree of the Polynomial of  $ 4{x^4} + 0{x^3} + 0{x^5} + 5x + 7 $ is

(A)  $ 4 $ 

(B)  $ 5 $ 

(C)  $ 3 $ 

(D)  $ 7 $ 

Ans: Option (A) is correct.

The highest power of x in the Polynomial is called the degree of Polynomial. We have the term with the highest power of x is  $ 4{x^4} $ , which is  $ 4 $ .

Therefore, the degree of Polynomial is  $ 4 $ .

4. Degree of the zero Polynomial

(A)  $ 0 $ 

(B)  $ 1 $ 

(C) Any natural number

(D) Not defined

Ans: Option (D) is the correct answer.

Like any constant value, the value 0 can be considered as a (constant) Polynomial, called the zero Polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. Therefore, the degree of the zero Polynomial is not defined.

5. If  $ p\left( x \right) = {x^2} - 2\sqrt 2 x + 1 $ , then  $ p\left( {2\sqrt 2 } \right) $  is equal to

(A)  $ 0 $  

(C)  $ 4\sqrt 2  $ 

(D)  $ 8\sqrt 2  + 1 $ 

Ans: The correct option is (B).

 $ p\left( x \right) = {x^2} - 2\sqrt 2 x + 1 $ 

Here, we have given the value of  $ x = 2\sqrt 2  $ .

 So, on putting the value in above equation, we get

 $ p\left( {2\sqrt 2 } \right) = {\left( {2\sqrt 2 } \right)^2} - 2\sqrt 2 \left( {2\sqrt 2 } \right) + 1 $ 

 $ p\left( {2\sqrt 2 } \right) = 4 \times 2 - 4 \times 2 + 1 $ 

 $ p\left( {2\sqrt 2 } \right) = 8 - 8 + 1 $ 

 $ p\left( {2\sqrt 2 } \right) = 1 $ 

6. The value of the Polynomial  $ 5x - 4{x^2} + 3 $ , when  $ x =  - 1 $  is

(A)  $  - 6 $ 

(B)  $ 6 $ 

(C)  $ 2 $ 

(D)  $  - 2 $ 

Ans: The correct option is (A).

Here we have given a Polynomial  $ 5x - 4{x^2} + 3 $ .

In this we have to put  $ x =  - 1 $ , and then we have to find the value of Polynomial

 $ p\left( x \right) = 5x - 4{x^2} + 3 $ 

 $ p\left( { - 1} \right) = 5\left( { - 1} \right) - 4{\left( { - 1} \right)^2} + 3 $ 

 $ p\left( { - 1} \right) =  - 5 - 4 + 3 $ 

 $ p\left( { - 1} \right) =  - 9 + 3 $ 

 $ p\left( { - 1} \right) =  - 6 $ 

7. If  $ p\left( x \right) = x + 3 $ , then  $ p\left( x \right) + p\left( { - x} \right) $  is equal to

(A)  $ 3 $ 

(B)  $ 2x $ 

(C)  $ 0 $ 

(D)  $ 6 $ 

Ans: The correct option is (D).

Here, we have  $ p\left( x \right) = x + 3 $ .

Now, on substituting  $ x $  with  $  - x $ , we get  $ p\left( { - x} \right) =  - x + 3 $ 

Now, on adding both the equations, we get

 $ p\left( x \right) + p\left( { - x} \right) = x + 3 - x + 3 $ 

 $ p\left( x \right) + p\left( { - x} \right) = 6 $ 

8. Zero of a zero Polynomial is 

(C) Any real number 

Ans: (C) is the correct answer.

Zero Polynomial is a constant Polynomial whose all coefficients are equal to 0. Zero of a Polynomial or we can say that root of a polynomial is the value of a variable that is responsible to make the Polynomial equals to zero. Hence, zero of the zero Polynomial is any real number.

9. Zero of the Polynomial  $ p\left( x \right) = 2x + 5 $  is

(A)  $  - \dfrac{2}{5} $ 

(B)  $  - \dfrac{5}{2} $ 

(C)  $ \dfrac{2}{5} $ 

(D)  $ \dfrac{5}{2} $ 

Ans : (B) is the correct option.

To get the zero of the Polynomial, we have to put the expression equals to zero.

 $ p\left( x \right) = 0 $ 

 $ 2x + 5 = 0 $ 

 $ 2x =  - 5 $ 

 $ x =  - \dfrac{5}{2} $ 

Therefore,  $  - \dfrac{5}{2} $  is the zero of the Polynomial.

10. One of the zeroes of given Polynomial  $ 2{x^2} + 7x - 4 $  is

(A)  $ 2 $ 

(B)  $ \dfrac{1}{2} $ 

(C)  $  - \dfrac{1}{2} $ 

(D)  $  - 2 $ 

Ans: (B) is the correct option.

To find the zeroes of a Polynomial, we have to put that expression equals to zero.

 $ 2{x^2} + 7x - 4 = 0 $ 

Using middle term splitting,

 $ 2{x^2} + \left( {8 - 1} \right)x - 4 = 0 $ 

 $ 2{x^2} + 8x - x - 4 = 0 $ 

 $ 2x\left( {x + 4} \right) - 1\left( {x + 4} \right) = 0 $ 

 $ \left( {2x - 1} \right)\left( {x + 4} \right) = 0 $ 

So, there we have two Ansutions

 $ 2x - 1 = 0 $  and  $ x + 4 = 0 $ 

 $ 2x = 1 $  and  $ x =  - 4 $ 

 $ x = \dfrac{1}{2} $  and  $ x =  - 4 $ 

11. If  $ {x^{51}} + 51 $  is divided by  $ x + 1 $ , the remainder is

(A)  $ 0 $ 

(B)  $ 1 $ 

(C)  $ 49 $ 

(D)  $ 50 $ 

Ans: Option (D) is the correct option.

Here we will use the remainder theorem.

If any Polynomial  $ f(x) $  is divided by  $ x - h $ , then the remainder will be  $ f\left( h \right) $ 

Here given Polynomial is  $ {x^{51}} + 51 $  and divided by  $ x + 1 $ 

So, by using the remainder theorem we can say that

Remainder=  $ f\left( { - 1} \right) $  since the divisor is  $ x + 1 $ 

Therefore we can write it as

\[ \Rightarrow f(x) = {x^{51}} + 51\]

\[ \Rightarrow f( - 1) = {( - 1)^{51}} + 51\]

\[ \Rightarrow f( - 1) =  - 1 + 51\]

\[\therefore f( - 1) = 50\]

12. If  $ x + 1 $ , is a factor of the Polynomial  $ 2{x^2} + kx $ , then the value of k is

(A)  $  - 3 $ 

(B)  $ 4 $ 

(C)  $ 2 $ 

Ans: Option (C) is correct.

Let  $ p\left( x \right) = 2{x^2} + kx $ 

Since, \[\left( {x{\text{ }} + {\text{ }}1} \right)\] is a factor of \[p\left( x \right)\] , then

Using, factor theorem

\[ \Rightarrow p\left( { - 1} \right) = 0\]

\[ \Rightarrow 2{\left( { - 1} \right)^2} + {\text{ }}k\left( { - 1} \right){\text{ }} = {\text{ }}0\]

\[ \Rightarrow \;2 - k{\text{ }} = {\text{ }}0\]

\[ \Rightarrow \;k = {\text{ }}2\]

Therefore, the value of k is  $ 2 $ .

13.  $ x + 1 $ , is a factor of the Polynomial

(A)  $ {x^3} + {x^2} - x + 1 $ 

(B)  $ {x^3} + {x^2} + x + 1 $ 

(C)  $ {x^4} + {x^3} + {x^2} + 1 $ 

(D)  $ {x^4} + 3{x^3} + 3{x^2} + x + 1 $ 

 If  $ x + 1 $  is the factor of the Polynomial, then  $ x =  - 1 $  is the root of the Polynomial.

(A) Let  $ p\left( x \right) = {x^3} + {x^2} - x + 1 $ 

Now, put  $ x =  - 1 $ .

 $ p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - \left( { - 1} \right) + 1 $ 

 $ p\left( { - 1} \right) =  - 1 + 1 + 1 + 1 $ 

 $ p\left( { - 1} \right) = 2 $ 

Therefore,  $ p\left( { - 1} \right) \ne 0 $ .

So,  $ x + 1 $  is not a factor of the  $ p\left( x \right) $ .

(B) Let  $ p\left( x \right) = {x^3} + {x^2} + x + 1 $ 

 $ p\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 $ 

 $ p\left( { - 1} \right) =  - 1 + 1 - 1 + 1 $ 

 $ p\left( { - 1} \right) = 0 $ 

Therefore,  $ p\left( { - 1} \right) = 0 $ .

So,  $ x + 1 $  is a factor of the  $ p\left( x \right) $ .

(C) Let  $ p\left( x \right) = {x^4} + {x^3} + {x^2} + 1 $ 

 $ p\left( { - 1} \right) = {\left( { - 1} \right)^4} + {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + 1 $ 

 $ p\left( { - 1} \right) = 1 - 1 + 1 + 1 $ 

(D) Let  $ p\left( x \right) = {x^4} + 3{x^3} + 3{x^2} + x + 1 $ 

 $ p\left( { - 1} \right) = {\left( { - 1} \right)^4} + 3{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 $ 

 $ p\left( { - 1} \right) = 1 - 3 + 3 - 1 + 1 $ 

 $ p\left( { - 1} \right) = 1 $ 

So,  $ x + 1 $  is not the factor of the  $ p\left( x \right) $ .

14. One of the factor of  $ \left( {25{x^2} - 1} \right) + {\left( {1 + 5x} \right)^2} $  is

(A)  $ 5 + x $ 

(B)  $ 5 - x $ 

(C)  $ 5x - 1 $ 

(D)  $ 10x $ 

Ans: (D) is the correct option.

 $  \Rightarrow \left( {25{x^2} - 1} \right) + {\left( {1 + 5x} \right)^2} $ 

 $  \Rightarrow \left( {{{\left( {5x} \right)}^2} - {{\left( 1 \right)}^2}} \right) + {\left( {5x + 1} \right)^2} $ 

 $  \Rightarrow \left( {5x + 1} \right)\left( {5x - 1} \right) + {\left( {5x + 1} \right)^2} $ 

Now, taking common  $ \left( {5x + 1} \right) $ 

 $  \Rightarrow \left( {5x + 1} \right)\left( {\left( {5x - 1} \right) + \left( {5x + 1} \right)} \right) $ 

 $  \Rightarrow \left( {5x + 1} \right)\left( {5x - 1 + 5x + 1} \right) $ 

 $  \Rightarrow \left( {5x + 1} \right)\left( {10x} \right) $ 

Therefore, one of the factor of  $ \left( {25{x^2} - 1} \right) + {\left( {1 + 5x} \right)^2} $  is  $ 10x $ .

15. The value of  $ {249^2} - {248^2} $  is 

(A)  $ {1^2} $ 

(B)  $ 477 $ 

(C)  $ 487 $ 

(D)  $ 497 $ 

Here we will use the identity  $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ .

We have, 

   $  \Rightarrow {249^2} - {248^2} $ 

 $  \Rightarrow \left( {249 - 248} \right)\left( {249 + 248} \right) $ 

 $  \Rightarrow \left( 1 \right)\left( {497} \right) $ 

 $  \Rightarrow 497 $ 

16. The factorization of  $ 4{x^2} + 8x + 3 $  is

(A)  $ \left( {x + 1} \right)\left( {x + 3} \right) $ 

(B)  $ \left( {2x + 1} \right)\left( {2x + 3} \right) $ 

(C)  $ \left( {2x + 2} \right)\left( {2x + 5} \right) $ 

(D)  $ \left( {2x - 1} \right)\left( {2x - 3} \right) $ 

In this question, we will use middle term splitting for factorization

 $  \Rightarrow 4{x^2} + 8x + 3 $ 

We can also write it as,

 $  \Rightarrow 4{x^2} + 6x + 2x + 3 $ 

 $  \Rightarrow 2x\left( {2x + 3} \right) + 1\left( {2x + 3} \right) $ 

 $  \Rightarrow \left( {2x + 1} \right)\left( {2x + 3} \right) $ 

17. Which of the followings is factor of  $ {\left( {x + y} \right)^3} - \left( {{x^3} + {y^3}} \right) $ ?

(A)  $ {x^2} + {y^2} + 2xy $ 

(B)  $ {x^2} + {y^2} - xy $ 

(C)  $ x{y^2} $ 

(D)  $ 3xy $ 

We have,  $ {\left( {x + y} \right)^3} - \left( {{x^3} + {y^3}} \right) $ .

Now, in this question we will use the identity  $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ 

 $  \Rightarrow {x^3} + {y^3} + 3xy\left( {x + y} \right) - \left( {{x^3} + {y^3}} \right) $ 

 $  \Rightarrow 3xy\left( {x + y} \right) $ 

So, we can say that  $ 3xy $  is a factor of  $ {\left( {x + y} \right)^3} - \left( {{x^3} + {y^3}} \right) $ .

18. The coefficient of  variable x in the expansion  $ {\left( {x + 3} \right)^3} $ is 

(B)  $ 9 $ 

(C)  $ 18 $ 

(D)  $ 27 $ 

Here in this question, we will use the identity  $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ 

Therefore, 

 $ {\left( {x + 3} \right)^3} = {x^3} + {3^3} + 3 \times x \times 3\left( {x + 3} \right) $ 

 $ {\left( {x + 3} \right)^3} = {x^3} + 27 + 9{x^2} + 27x $ 

Therefore, the coefficient of x is  $ 27 $ .

19. If  $ \dfrac{x}{y} + \dfrac{y}{x} =  - 1 $  the value of  $ {x^3} - {y^3} $  is

Ans: The correct option is (C).

Here, it is given that   $ \dfrac{x}{y} + \dfrac{y}{x} =  - 1 $ 

On taking LCM, we get

 $ \dfrac{{{x^2} + {y^2}}}{{xy}} =  - 1 $ 

On cross-multiplication, we get

 $ {x^2} + {y^2} =  - xy $ 

Now, we know that

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + {y^2} + xy} \right) $ 

On putting  $ {x^2} + {y^2} =  - xy $  in above equation, we get

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( { - xy + xy} \right) $ 

 $ {x^3} - {y^3} = \left( {x - y} \right)\left( 0 \right) $ 

 $ {x^3} - {y^3} = 0 $ 

20. If  $ 49{x^2} - b = \left( {7x + \dfrac{1}{2}} \right)\left( {7x - \dfrac{1}{2}} \right) $  , then the value of b is

(A)   $ 0 $ 

(B)  $ \dfrac{1}{{\sqrt 2 }} $ 

(C)  $ \dfrac{1}{4} $ 

 Here in this question, we will use the identity  $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ 

 $  \Rightarrow 49{x^2} - b = \left( {7x + \dfrac{1}{2}} \right)\left( {7x - \dfrac{1}{2}} \right) $ 

 $  \Rightarrow 49{x^2} - b = {\left( {7x} \right)^2} - {\left( {\dfrac{1}{2}} \right)^2} $ 

 $  \Rightarrow 49{x^2} - b = 49{x^2} - {\left( {\dfrac{1}{2}} \right)^2} $ 

On comparing, we get 

 $ b = {\left( {\dfrac{1}{2}} \right)^2} $ 

 $ b = \dfrac{1}{4} $ 

21. If  $ a + b + c = 0 $ ,  then the value of  $ {a^3} + {b^3} + {c^3} $  is equal to

(B)  $ abc $ 

(C)  $ 3abc $ 

(D)  $ 2abc $ 

Here we know an identity, 

 $ {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $ 

It is given that, $ a + b + c = 0 $ 

Therefore, put it in the above equation

 $  \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $ 

 $  \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 $ 

 $  \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $ 

Short Answer Questions with Reasoning

Sample Question 1: Write whether the following statements are True or False.

Justify your answer .

(i)  $ \dfrac{1}{{\sqrt 5 }}{x^{\dfrac{1}{2}}} + 1 $  is a polynomial

Ans: It is a false statement.

Here, the exponent of a variable is not a whole number. So, it is not a polynomial.

(ii)  $ \dfrac{{6\sqrt x  + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} $  is a polynomial,  $ x \ne 0 $ 

Ans: It is a true statement.

We can also write it as, 

 $ \dfrac{{6\sqrt x  + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} = \dfrac{{\sqrt x {{\left( {6 + x} \right)}^{}}}}{{\sqrt x }} = 6 + x $ 

Here, the exponent of variable is not a whole number. So, it is a polynomial.

EXERCISE 2.2

1. Which of the following expressions are Polynomials? Justify your answer.

(i)  $ 8 $ 

Ans:  Here we have given an expression  $ 8 $ in which the degree of variable is zero. Therefore, it is a constant Polynomial.

(ii)  $ \sqrt 3 {x^2} - 2x $ 

Ans: $ \sqrt 3 {x^2} - 2x $ 

Here, in each term the power of x is a whole number. Therefore, this expression is a Polynomial.

(iii)  $ 1 - \sqrt {5x}  $ 

Ans: $ 1 - \sqrt {5x}  $ 

We can also write it as  $ 1 - \sqrt 5 {x^{\dfrac{1}{2}}} $ . Here the degree of variable is not a whole number. Therefore, this expression is not a Polynomial.

(iv)  $ \dfrac{1}{{5{x^{ - 2}}}} + 5x + 7 $ 

Ans: $ \dfrac{1}{{5{x^{ - 2}}}} + 5x + 7 $ 

We can also write it as  $ \dfrac{{1{x^2}}}{5} + 5x + 7 $ . Here in this expression each variable term has the power of x in a whole number. Therefore, this expression is a Polynomial.

(v)  $ \dfrac{{\left( {x - 2} \right)\left( {x - 4} \right)}}{x} $ 

Ans: $ \dfrac{{\left( {x - 2} \right)\left( {x - 4} \right)}}{x} $ 

We can also write it as 

 $ \dfrac{{{x^2} - 6x + 8}}{x} = x - 6 + \dfrac{8}{x} = x - 6 + 8{x^{ - 1}} $ 

Here, the exponent of variable x in the third term  $ 8{x^{ - 1}} $  is – 1, and this is not a whole number.

 Therefore, this algebraic expression is not a Polynomial.

(vi)  $ \dfrac{1}{{x + 1}} $

Ans: $ \dfrac{1}{{x + 1}} $ 

We can also reduce it to  $ {\left( {x + 1} \right)^{ - 1}} $ , which cannot be reduced to an expression in which the exponent of the variable x had only whole numbers in each of its terms. Therefore, this algebraic expression is not a Polynomial.

(vii)  $ \dfrac{1}{7}{a^3} - \dfrac{2}{{\sqrt 3 }}{a^2} + 4a - 7 $ 

Ans: $ \dfrac{1}{7}{a^3} - \dfrac{2}{{\sqrt 3 }}{a^2} + 4a - 7 $ 

In this expression, the exponent of a in each term is a whole number, so this expression is a Polynomial.

(viii)  $ \dfrac{1}{{2x}} $ 

Ans: $ \dfrac{1}{{2x}} $ 

We can also write it as  $ \dfrac{1}{2}{x^{ - 1}} $ . Here, the exponent of the variable x is   $  - 1 $  , which is not a whole number so, this algebraic expression is not a Polynomial.

2. Write whether the following statements are True or False. Justify your answer. 

(i) A binomial can have at most two terms 

Ans: The given statement is False because binomials have exactly two terms.

(ii) Every Polynomial is a binomial 

Ans: Every Polynomial cannot be a binomial because a Polynomial had many terms but a binomial had only two terms in the expression. For example,  $ {x^4} + {x^3} + {x^2} + x + 1 $ is a Polynomial but not a binomial. Therefore, the given statement is false.

(iii) A binomial may have degree  $ 5 $ 

Ans: The given statement is True because a binomial is a Polynomial whose degree is a whole number and it is greater than or equals to one and contain two terms. For example,  $ {x^5} - 1 $  is a binomial of degree  $ 5 $ .

(iv) Zero of a Polynomial is always  $ 0 $  

Ans: The given statement is False, because zero of Polynomial can be any real number.

(v) A Polynomial can't have more than one zeros 

Ans: The given statement is False, because a Polynomial can have any number of zeroes which depends on the degree of the Polynomial.

(vi) The power of the sum of two polynomials each of whose degree  $ 5 $  is always  $ 5 $ .

Ans: The given statement is False. For example, consider the two Polynomials  $  - {x^5} + 3{x^2} + 4 $  and  $ {x^5} + {x^4} + 2{x^3} + 3 $ .The degree of each of these polynomials will be 5. Their sum is  $ {x^4} + 2{x^3} + 3{x^2} + 7 $ . The degree of this Polynomial is  $ 4 $  not  $ 5 $ .

Short Answer Questions

Sample Question 1: (i) Check whether p(x) is a multiple of g(x) or not, where \[\;p\left( x \right) = {x^3}-{\text{ }}x + 1,\,{\text{ }}g\left( x \right) = 2-3x\]

Ans: p(x) will be a multiple of g(x) if and only if g(x) divides p(x).  

Now,  $ g\left( x \right) = 2 - 3x $ 

But,  $ g\left( x \right) = 0 $ 

 $  \Rightarrow 2 - 3x = 0 $ 

 $  \Rightarrow x = \dfrac{2}{3} $ 

Now, put  $ x = \dfrac{2}{3} $ in equation \[\;p\left( x \right) = {x^3}-{\text{ }}x + 1\]

\[\;p\left( {\dfrac{2}{3}} \right) = {\left( {\dfrac{2}{3}} \right)^3}-{\text{ }}\left( {\dfrac{2}{3}} \right) + 1\]

\[\;p\left( {\dfrac{2}{3}} \right) = \dfrac{8}{{27}} + \dfrac{1}{3} = \dfrac{{8 + 9}}{{27}} = \dfrac{{17}}{{27}}\]

Here, the remainder is not equals to zero.

Therefore, g(x) is not a multiple of p(x).

(ii) Check whether g(x) is a factor of p(x) or not, where \[p\left( x \right) = {\text{ }}8{x^3}-6{x^2}-4x + 3,{\text{ }}g\left( x \right) = {\text{ }}\dfrac{x}{3} - \dfrac{1}{4}\]

Now,  $ g\left( x \right) = \dfrac{x}{3} - \dfrac{1}{4} $ 

 $  \Rightarrow \dfrac{x}{3} - \dfrac{1}{4} = 0 $ 

 $  \Rightarrow x = \dfrac{3}{4} $ 

Now, put  $ x = \dfrac{3}{4} $ in equation \[p\left( x \right) = {\text{ }}8{x^3}-6{x^2}-4x + 3\]

\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}8{\left( {\dfrac{3}{4}} \right)^3}-6{\left( {\dfrac{3}{4}} \right)^2}-4\left( {\dfrac{3}{4}} \right) + 3\]

\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}8\left( {\dfrac{{27}}{{64}}} \right)-6\left( {\dfrac{9}{{16}}} \right)-3 + 3\]

\[p\left( {\dfrac{3}{4}} \right) = {\text{ }}\left( {\dfrac{{27}}{8}} \right)-\left( {\dfrac{{27}}{8}} \right)-3 + 3 = 0\]

Here, the remainder is equals to zero.

Therefore, g(x) is a multiple of p(x).

Sample Question 2: Find the value of a, if \[x-a\] is a factor of \[{x^3}-a{x^2}{\text{ }} + {\text{ }}2x{\text{ }} + {\text{ }}a{\text{ }}-{\text{ }}1\].

Ans: Here, \[p\left( x \right) = {x^3}-a{x^2}{\text{ }} + {\text{ }}2x{\text{ }} + {\text{ }}a{\text{ }}-{\text{ }}1\]

If \[x-a\] is a factor of p(x). Then,  $ p\left( a \right) = 0 $ 

\[ \Rightarrow {a^3}-a{\left( a \right)^2}{\text{ }} + {\text{ }}2\left( a \right){\text{ }} + {\text{ }}a{\text{ }}-{\text{ }}1 = 0\]

 $  \Rightarrow 3a - 1 = 0 $ 

 $  \Rightarrow a = \dfrac{1}{3} $ 

Therefore,  $ a = \dfrac{1}{3} $ .

Sample Question 3: (i) Without actually calculating the cubes, find the value of \[{48^3}{\text{ }}-{\text{ }}{30^3}{\text{ }}-{\text{ }}{18^3}\].

Ans: We know that, \[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} + {\text{ }}{z^3}{\text{ }}-{\text{ }}3xyz{\text{ }} = {\text{ }}\left( {x{\text{ }} + {\text{ }}y{\text{ }} + {\text{ }}z} \right){\text{ }}\left( {{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} + {\text{ }}{z^2}{\text{ }}-{\text{ }}xy{\text{ }}-{\text{ }}yz{\text{ }}-{\text{ }}zx} \right)\]

If,  $ x + y + z = 0 $ 

Then, \[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} + {\text{ }}{z^3}{\text{  =  }}3xyz{\text{ }}\]

 Here,  $ 48 - 30 - 18 = 0 $ 

Therefore,   \[{48^3}{\text{ }}-{\text{ }}{30^3}{\text{ }}-{\text{ }}{18^3} = 3\left( {48} \right)\left( { - 30} \right)\left( { - 18} \right) = 77760\]

(ii)Without finding the cubes, factorise \[{\left( {x{\text{ }}-{\text{ }}y} \right)^3}{\text{ }} + {\text{ }}{\left( {y{\text{ }}-{\text{ }}z} \right)^3}{\text{ }} + {\text{ }}{\left( {z{\text{ }}-{\text{ }}x} \right)^3}\] .

Here,  $ \left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 $ 

Therefore,   \[{\left( {x{\text{ }}-{\text{ }}y} \right)^3}{\text{ }} + {\text{ }}{\left( {y{\text{ }}-{\text{ }}z} \right)^3}{\text{ }} + {\text{ }}{\left( {z{\text{ }}-{\text{ }}x} \right)^3} = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\]

EXERCISE 2.3

1. Classify the following Polynomial as Polynomials in one variable, two variables etc.

(i)  $ {x^2} + x + 1 $ 

Ans: $ {x^2} + x + 1 $  is a Polynomial in one variable. 

(ii)  $ {y^3} - 5y $ 

Ans: $ {y^3} - 5y $  is a Polynomial in one variable.

(iii)  $ xy + yz + zx $ 

Ans: $ xy + yz + zx $  is a Polynomial in three variables.

(iv)  $ {x^2} - 2xy + {y^2} + 1 $ 

Ans: $ {x^2} - 2xy + {y^2} + 1 $  is a Polynomial in two variables.

2. Determine the degree of each of the following Polynomials:

(i)  $ 2x - 1 $ 

Ans: Since the highest power of x is  $ 1 $ , the degree of the Polynomial  $ 2x - 1 $  is  $ 1 $ .

(ii) $  - 10 $ 

Ans: $  - 10 $  is a non-zero constant. A constant term does not contain any variable and its degree is always 0.

(iii) $ {x^3} - 9x + 3{x^5} $ 

Ans: Since the highest power of x is  $ 5 $ , the degree of the Polynomial  $ {x^3} - 9x + 3{x^5} $  is  $ 5 $ .

(iv) $ {y^3}\left( {1 - {y^4}} \right) $ 

Ans: Here, we have $ {y^3}\left( {1 - {y^4}} \right) = {y^3} - {y^7} $ . Since the highest power of y is  $ 7 $ , the degree of the Polynomial is  $ 7 $ .

3. For the Polynomial  $ \dfrac{{{x^3} + 2x + 1}}{5} - \dfrac{7}{2}{x^2} - {x^6} $, write 

(i) the degree of the Polynomial 

Ans: $ \dfrac{{{x^3} + 2x + 1}}{5} - \dfrac{7}{2}{x^2} - {x^6} $ 

 $ \dfrac{{{x^3}}}{5} + \dfrac{2}{5}x + \dfrac{1}{5} - \dfrac{7}{2}{x^2} - {x^6} $ 

As we know that highest power of variable in a Polynomial is known as  degree of a Polynomial. In given Polynomial, the term with highest of x is  $  - {x^6} $ , and the exponent of x in this term in  $ 6 $ .

(ii) the coefficient of  $ {x^3} $ 

Ans: The coefficient of  $ {x^3} $ is  $ \dfrac{1}{5} $ .

(iii) the coefficient of  $ {x^6} $ 

Ans:  The coefficient of  $ {x^6} $  is  $  - 1 $ .

(iv) the constant term.

Ans: The constant term is  $ \dfrac{1}{5} $ .

4. Write the coefficient of  $ {x^2} $  in each of the following:

(i)  $ \dfrac{\pi }{6}x + {x^2} - 1 $ 

Ans: Coefficient of  $ {x^2} $  in the given Polynomial is  $ 1 $ .

(ii)  $ 3x - 5 $ 

Ans: The given Polynomial can also be written as  $ 0{x^2} + 3x - 5 $ . So, the coefficient of  $ {x^2} $  in the Polynomial is  $ 0 $ .

(iii)  $ \left( {x - 1} \right)\left( {3x - 4} \right) $ 

Ans: We can write the given polynomial as:

 $  \Rightarrow \left( {x - 1} \right)\left( {3x - 4} \right) $ 

 $  \Rightarrow 3{x^2} - 3x - 4x + 4 $ 

 $  \Rightarrow 3{x^2} - 7x + 4 $ 

Therefore, the coefficient of  $ {x^2} $  in the given Polynomial is  $ 3 $ .

(iv)  $ \left( {2x - 5} \right)\left( {2{x^2} - 3x + 1} \right) $ 

 $  \Rightarrow \left( {2x - 5} \right)\left( {2{x^2} - 3x + 1} \right) $ 

 $  \Rightarrow 4{x^3} - 6{x^2} + 2x - 10{x^2} + 15x - 5 $ 

 $  \Rightarrow 4{x^3} - 16{x^2} + 17x - 5 $ 

So, the coefficient of  $ {x^2} $  in the given Polynomial is – 16.

5. Classify the following as a constant, linear quadratic and cubic Polynomials:

(i)  $ 2 - {x^2} - {x^3} $ 

Ans: To do this question, we have to keep some points in mind:

(A) A Polynomial in which there is no variable term and there is only constant term, is known as a constant polynomial. 

(B) A Polynomial having degree 1 is known as a linear Polynomial. 

(C) A Polynomial of having degree 2 is known as quadratic Polynomial. 

(D) A Polynomial having degree 3 is known as cubic Polynomial. 

Cubic Polynomial is the correct answer.

(ii)  $ 3{x^3} $ 

Ans: Cubic Polynomial is the correct answer.

(iii)  $ 5t - \sqrt 7  $ 

Ans: Linear Polynomial is the correct answer.

(iv)  $ 4 - 5{y^2} $ 

Ans: Quadratic Polynomial is the correct answer.

(v)  $ 3 $ 

Ans: Constant Polynomial is the correct answer.

(vi)  $ 2 + x $ 

(vii)  $ {y^3} - y $ 

(viii)  $ 1 + x + {x^3} $ 

Ans:  Cubic Polynomial is the correct answer.

(ix)  $ {t^2} $ 

(x)  $ \sqrt 2 x - 1 $ 

6. Give an example of a Polynomial, which is:

(i) monomial of degree  $ 1 $  

Ans: A Polynomial which contains  only one term is called a monomial, a Polynomial having only two terms is called binomial, a Polynomial having only three terms is called a trinomial.

$ 5x $  is monomial of degree  $ 1 $ .

(ii) binomial of degree  $ 20 $  

Ans: $ {x^{20}} + 5 $  is a binomial of degree  $ 20 $ . 

(iii) trinomial of degree  $ 2 $ 

Ans: $2{x^2} + 2x + 1 $  is a trinomial of degree  $ 2 $ .

7. Find the value of the Polynomial  $ 3{x^3} - 4{x^2} + 7x + 5 $ , when  $ x = 3 $  and also when  $ x =  - 3 $ .

Ans: Let p(x) be the given Polynomial.

 $ p\left( x \right) = 3{x^2} - 4{x^2} + 7x - 5 $ 

Now, put  $ x = 3 $ 

 $ p\left( 3 \right) = 3{\left( 3 \right)^3} - 4{\left( 3 \right)^2} + 7\left( 3 \right) - 5 $ 

 $ p\left( 3 \right) = 3\left( {27} \right) - 4\left( 9 \right) + 7\left( 3 \right) - 5 $ 

 $ p\left( 3 \right) = 61 $ 

Put  $ x =  - 3 $ 

 $ p\left( { - 3} \right) = 3{\left( { - 3} \right)^3} - 4{\left( { - 3} \right)^2} + 7\left( { - 3} \right) - 5 $ 

 $ p\left( { - 3} \right) = 3\left( { - 27} \right) - 4\left( 9 \right) + 7\left( { - 3} \right) - 5 $ 

 $ p\left( { - 3} \right) =  - 81 - 36 - 21 - 5 $ 

 $ p\left( { - 3} \right) =  - 143 $ 

8. If  $ p\left( x \right) = {x^2} - 4x + 3 $ , evaluate  $ p\left( 2 \right) - p\left( { - 1} \right) + p\left( {\dfrac{1}{2}} \right) $ 

Ans: We have  $ p\left( x \right) = {x^2} - 4x + 3 $ 

Now, put  $ x = 2 $ 

 $ p\left( 2 \right) = \left( {{2^2} - 4 \times 2 + 3} \right) = \left( {4 - 8 + 3} \right) = \left( { - 4 + 3} \right) =  - 1 $ 

Now, put  $ x =  - 1 $ 

 $ p\left( { - 1} \right) = \left( {{{\left( { - 1} \right)}^2} - 4 \times \left( { - 1} \right) + 3} \right) = \left( {1 + 4 + 3} \right) = 8 $ 

Now, put  $ x = \dfrac{1}{2} $ 

 $ p\left( {\dfrac{1}{2}} \right) = \left( {{{\left( {\dfrac{1}{2}} \right)}^2} - 4 \times \dfrac{1}{2} + 3} \right) = \left( {\dfrac{1}{4} - 2 + 3} \right) = \left( {1 + \dfrac{1}{4}} \right) = \dfrac{5}{4} $ 

 $ p\left( 2 \right) - p\left( { - 1} \right) + p\left( {\dfrac{1}{2}} \right) =  - 1 - 8 + \dfrac{5}{4} $ 

 $ p\left( 2 \right) - p\left( { - 1} \right) + p\left( {\dfrac{1}{2}} \right) =  - 9 + \dfrac{5}{4} = \dfrac{{ - 36 + 5}}{4} =  - \dfrac{{31}}{4} $ 

9. Find  $ p\left( 0 \right) $  ,  $ p\left( 1 \right) $  ,  $ p\left( { - 2} \right) $  for the following Polynomials:

(i)  $ p\left( x \right) = 10x - 4{x^2} - 3 $

Ans: We have

 $ p\left( x \right) = 10x - 4{x^2} - 3 $ 

Put,  $ x = 0 $ 

 $ p\left( 0 \right) = 10\left( 0 \right) - 4{\left( 0 \right)^2} - 3 =  - 3 $ 

Put,  $ x = 1 $ 

 $ p\left( 1 \right) = 10\left( 1 \right) - 4{\left( 1 \right)^2} - 3 = 10 - 4 - 3 = 3 $ 

Put,  $ x =  - 2 $ 

 $ p\left( { - 2} \right) = 10\left( { - 2} \right) - 4{\left( { - 2} \right)^2} - 3 =  - 20 - 16 - 3 =  - 39 $ 

(ii)  $ p\left( y \right) = \left( {y + 2} \right)\left( {y - 2} \right) $ 

 $ p\left( y \right) = \left( {y + 2} \right)\left( {y - 2} \right) $ 

Put,  $ y = 0 $ 

 $ p\left( 0 \right) = \left( {0 + 2} \right)\left( {0 - 2} \right) =  - 4 $ 

Put,  $ y = 1 $ 

 $ p\left( 1 \right) = \left( {1 + 2} \right)\left( {1 - 2} \right) = \left( 3 \right)\left( { - 1} \right) =  - 3 $ 

Put,  $ y =  - 2 $ 

 $ p\left( { - 2} \right) = \left( { - 2 + 2} \right)\left( { - 2 - 2} \right) = 0 $ 

10. Verify whether the following are true or false. 

(i)  $  - 3 $  is a zero of  $ x - 3 $ . 

Ans: A number c can be a zero of a Polynomial p(x) if a number c is such that \[p\left( c \right){\text{ }} = {\text{ }}0\].

 Let  $ p\left( x \right) = x - 3 $ 

 $ p\left( { - 3} \right) =  - 3 - 3 =  - 6 $ 

 $ p\left( { - 3} \right) \ne 0 $ 

Hence,  $  - 3 $  is not a zero of  $ x - 3 $  .

So, the result is False.

(ii)  $  - \dfrac{1}{3} $  is a zero of  $ 3x + 1 $ . 

Let  $ p\left( x \right) = 3x + 1 $ 

Put,  $ x =  - \dfrac{1}{3} $ 

 $ p\left( { - \dfrac{1}{3}} \right) = 3\left( { - \dfrac{1}{3}} \right) + 1 $ 

 $ p\left( { - \dfrac{1}{3}} \right) =  - 1 + 1 = 0 $ 

Hence,  $  - \dfrac{1}{3} $  is zero of  $ p\left( x \right) = 3x + 1 $ .

So, the result is True.

(iii)  $  - \dfrac{4}{5} $  is a zero of  $ 4 - 5y $ . 

Let  $ p\left( y \right) = 4 - 5(y) $ 

Put,  $ y =  - \dfrac{4}{5} $ 

 $ p\left( { - \dfrac{4}{5}} \right) = 4 - 5\left( { - \dfrac{4}{5}} \right) $ 

 $ p\left( { - \dfrac{4}{5}} \right) = 4 + 4 = 8 $ 

Hence,  $  - \dfrac{4}{5} $  is zero of  $ p\left( y \right) = 4 - 5(y) $ .

(iv)  $ 0 $  and  $ 2 $  are the zeroes of  $ {t^2} - 2t $ 

Ans: Let  $ p\left( t \right) = {t^2} - 2t $ 

Put, $ t = 0 $ 

 $ p\left( 0 \right) = {\left( 0 \right)^2} - 2\left( 0 \right) $ 

 $ p\left( 0 \right) = 0 $ 

Put, $ t = 2 $ 

 $ p\left( 2 \right) = {\left( 2 \right)^2} - 2\left( 2 \right) = 4 - 4 = 0 $ 

Hence,  $ 0 $  and  $ 2 $  are zeroes of the Polynomial  $ {t^2} - 2t $ .

(v)   $  - 3 $  is a zero  $ {y^2} + y - 6 $ 

Ans: Let  $ p\left( y \right) = {y^2} + y - 6 $ 

Put,  $ y =  - 3 $ 

 $ p\left( { - 3} \right) = {\left( { - 3} \right)^2} + \left( { - 3} \right) - 6 $ 

 $ p\left( { - 3} \right) = 9 - 9 = 0 $ 

Hence,  $  - 3 $   is a zero of the Polynomial  $ {y^2} + y - 6 $ .

11.  Find the zeroes of the Polynomial in each of the following:

(i)  $ p\left( x \right) = x - 4 $ 

Ans: Here we have to Solve the equation  $ p\left( x \right) = 0 $ , we get

 $ x - 4 = 0 $ 

 $ x = 4 $  

So,  $ 4 $  is a zero of the Polynomial  $ x - 4 $ .

(ii)  $ g\left( x \right) = 3 - 6x $ 

Ans: Here we have to Solve the equation  $ g\left( x \right) = 0 $ , we get

 $ 3 - 6x = 0 $ 

 $ x = \dfrac{1}{2} $  

So,  $ \dfrac{1}{2} $  is a zero of the Polynomial  $ 3 - 6x $ .

(iii)  $ q\left( x \right) = 2x - 7 $ 

Ans: Here we have to Solve the equation  $ q\left( x \right) = 0 $ , we get

 $ 2x - 7 = 0 $ 

 $ x = \dfrac{7}{2} $  

So,  $ \dfrac{7}{2} $  is a zero of the Polynomial  $ 2x - 7 $ .

(iv)  $ h\left( y \right) = 2y $ 

Ans: Here we have to Solve the equation  $ h\left( y \right) = 0 $ , we get

 $ 2y = 0 $ 

 $ y = 0 $  

So,  $ 0 $  is a zero of the Polynomial  $ 2y $ .

12. Find the zeroes of the Polynomial  $ {\left( {x - 2} \right)^2} - {\left( {x + 2} \right)^2} $ .

Ans: Let  $ p\left( x \right) = {\left( {x - 2} \right)^2} - {\left( {x + 2} \right)^2} $ 

To get the zeroes of p(x), we have to Solve the equation \[p\left( x \right){\text{ }} = {\text{ }}0\].

So, \[p\left( x \right){\text{ }} = {\text{ }}0\]

 $ {\left( {x - 2} \right)^2} - {\left( {x + 2} \right)^2} = 0 $ 

 $ \left( {\left( {x - 2} \right) - \left( {x + 2} \right)} \right)\left( {\left( {x - 2} \right) + \left( {x + 2} \right)} \right) = 0 $ 

\[\left( { - 4} \right)\left( {2x} \right) = 0\]

 $ x = 0 $ 

Hence,  $ x = 0 $  is the only one zero of  $ p\left( x \right) $ .

13. By acute division, find the quotient and the remainder when the first Polynomial is divided by the second  $ {x^4} + 1\,;\,x + 1 $ .

Ans: By acute division, we have

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14. By remainder Theorem find the remainder, when p(x) is divided by g(x), where

(i)  $ p\left( x \right) = {x^3} - 2{x^2} - 4x - 1\,,\,g\left( x \right) = x + 1 $ 

Ans:  According to the remainder theorem, if we divide a polynomial  p(x)  by a factor  \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}-{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] ; then you will find a smaller polynomial with a remainder. This remainder which has been obtained is actually a value of  p(x) at  $ x = a $ , specifically p(a) . So, \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}-{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] is the divisor of p(x) if and only if \[p\left( a \right){\text{ }} = {\text{ }}0\] .

We have,  $ g\left( x \right) = x + 1 $ 

Now, put 

\[g\left( x \right) = 0\]

 $  \Rightarrow x + 1 = 0 $ 

  $  \Rightarrow x =  - 1 $ 

Remainder  $  = \,p\left( { - 1} \right) $ 

 $  = {\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^2} - 4\left( { - 1} \right) =  - 1 - 2 + 4 - 1 = 0 $ 

(ii)  $ p\left( x \right) = {x^3} - 3{x^2} + 4x + 50\,,\,g\left( x \right) = x - 3 $ 

Ans: According to the remainder theorem, if we divide a polynomial  p(x)  by a factor  \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}-{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] ; then you will find a smaller polynomial with a remainder. This remainder which has been obtained is actually a value of  p(x) at  $ x = a $ , specifically p(a) . So, \[\left( {{\text{ }}{\mathbf{x}}{\text{ }}-{\text{ }}{\mathbf{a}}{\text{ }}} \right)\] is the divisor of p(x) if and only if \[p\left( a \right){\text{ }} = {\text{ }}0\] .

We have,  $ g\left( x \right) = x - 3 $ 

 $  \Rightarrow x - 3 = 0 $ 

  $  \Rightarrow x = 3 $ 

Remainder  $  = \,p\left( 3 \right) $ 

 $  = {\left( 3 \right)^3} - 3{\left( 3 \right)^2} + 4\left( 3 \right) + 50 = 27 - 27 + 12 + 50 = 62 $ 

(iii)  $ p\left( x \right) = 4{x^3} - 12{x^2} + 14x - 3\,,\,g\left( x \right) = 2x - 1 $ 

We have,  $ g\left( x \right) = 2x - 1 $ 

 $  \Rightarrow 2x - 1 = 0 $ 

  $  \Rightarrow x = \dfrac{1}{2} $ 

Remainder  $  = \,p\left( {\dfrac{1}{2}} \right) $ 

 $  = 4{\left( {\dfrac{1}{2}} \right)^3} - 12{\left( {\dfrac{1}{2}} \right)^2} + 14\left( {\dfrac{1}{2}} \right) - 3 = 4\left( {\dfrac{1}{8}} \right) - 12\left( {\dfrac{1}{4}} \right) + 7 - 3 $ 

 $  = \dfrac{1}{2} - 3 + 7 - 3 = \dfrac{1}{2} + 1 = \dfrac{3}{2} $ 

(iv)  $ p\left( x \right) = {x^3} - 6{x^2} + 2x - 4\,,\,g\left( x \right) = 1 - \dfrac{3}{2}x $ 

 We have,  $ g\left( x \right) = 1 - \dfrac{3}{2}x $ 

 $  \Rightarrow 1 - \dfrac{3}{2}x = 0 $ 

  $  \Rightarrow x = \dfrac{2}{3} $ 

Remainder  $  = \,p\left( {\dfrac{2}{3}} \right) $ 

 $  = {\left( {\dfrac{2}{3}} \right)^3} - 6{\left( {\dfrac{2}{3}} \right)^2} + 2\left( {\dfrac{2}{3}} \right) - 4\, = \dfrac{8}{{27}} - 6 \times \dfrac{4}{9} + \dfrac{4}{3} - 4 $ 

 $  = \dfrac{8}{{27}} - \dfrac{8}{3} + \dfrac{4}{3} - 4 = \dfrac{{8 - 72 + 36 - 108}}{{27}} = \dfrac{{ - 136}}{{27}} $ 

15. Check whether p(x) is a multiple of g(x) or not:

(i)  $ p\left( x \right) = {x^3} - 5{x^2} + 4x - 3\,,\,g\left( x \right) = x - 2 $ 

p(x) is a multiple of g(x) if g(x) divides p(x)

Now,  $ g\left( x \right) = x - 2 $  gives  $ x = 2 $ if we put  $ g\left( x \right) = 0 $ .

Remainder  $  = \,p\left( 2 \right) = {\left( 2 \right)^3} - 5{\left( 2 \right)^2} + 4\left( 2 \right) - 3 $ 

 $  = 8 - 5\left( 4 \right) + 8 - 3 = 8 - 20 + 8 - 3 =  - 7 $ 

Hence, the remainder  $  \ne 0 $ 

Therefore, function p(x) is not a multiple of function g(x).

(ii)  $ p\left( x \right) = 2{x^3} - 11{x^2} - 4x + 5\,,\,g\left( x \right) = 2x + 1 $ 

p(x) will be a multiple of g(x) if and only if g(x) divides p(x).

 $ g\left( x \right) = 2x + 1 $  give  $ x =  - \dfrac{1}{2} $ 

remainder  $  = p\left( { - \dfrac{1}{2}} \right) = 2{\left( {\dfrac{{ - 1}}{2}} \right)^3} - 11{\left( { - \dfrac{1}{2}} \right)^2} - 4\left( { - \dfrac{1}{2}} \right) + 5 $ 

 $  = 2\left( { - \dfrac{1}{8}} \right) - 11\left( {\dfrac{1}{4}} \right) + 2 + 5 = \dfrac{{ - 1}}{4} - \dfrac{{11}}{4} + 7 $ 

 $  = \dfrac{{ - 1 - 11 + 28}}{4} = \dfrac{{16}}{4} = 4 $ 

Since the remainder  $  \ne 0 $ .

So, p(x) will not be a multiple of g(x).

16. Show that:

(i)  $ x + 3 $  is a factor of  $ 69 + 11x - {x^2} + {x^3} $ 

Ans: Let  $ p\left( x \right) = 69 + 11x - {x^2} + {x^3}\,\,,\,\,g\left( x \right) = x + 3 $ 

 $ g\left( x \right) = x + 3 = 0 $ 

It will give,  $ x =  - 3 $ 

g(x) will be a factor of p(x) if  $ p\left( { - 3} \right) = 0 $  (Factor theorem).

Now,  $ p\left( { - 3} \right) = 69 + 11\left( { - 3} \right) - {\left( { - 3} \right)^2} + {\left( { - 3} \right)^3} $ 

 $  = 69 - 33 - 9 - 27 = 0 $ 

Hence  $ p\left( { - 3} \right) = 0 $ ,  g(x) will be a factor of p(x).

(ii)  $ 2x - 3 $  is a factor of  $ x + 2{x^3} - 9{x^2} + 12 $ 

Ans: Let  $ p\left( x \right) = x + 2{x^3} - 9{x^2} + 12 $  and  $ g\left( x \right) = 2x - 3 $ 

 $ g\left( x \right) = 2x - 3 $  

 $ g\left( x \right) = 0 $ 

 $ 2x - 3 = 0 $ 

 $ x = \dfrac{3}{2} $ , 

g(x) will be factor of p(x) if  $ p\left( {\dfrac{3}{2}} \right) = 0 $  (Factor theorem)

Now,  $ p\left( {\dfrac{3}{2}} \right) = \dfrac{3}{2} + 2{\left( {\dfrac{3}{2}} \right)^3} - 9{\left( {\dfrac{3}{2}} \right)^2} + 12 = \dfrac{3}{2} + 2\left( {\dfrac{{27}}{8}} \right) - 9\left( {\dfrac{9}{4}} \right) + 12 $ 

 $  = \dfrac{3}{2} + \dfrac{{27}}{4} - \dfrac{{81}}{4} + 12 = \dfrac{{6 + 27 - 81 + 48}}{4} = \dfrac{0}{4} = 0 $ 

Since,  $ p\left( {\dfrac{3}{2}} \right) = 0 $ , therefore, g(x) is a factor of p(x).

17. Determine which of the following Polynomials has  $ x - 2 $ the factor as:

(i)  $ 3{x^2} + 6x - 24 $ 

Ans: As we know that if  $ \left( {x - a} \right) $ is a factor of p(x), then p(A) =0.

Let  $ p\left( x \right) = 3{x^2} + 6x - 24 $ 

If  $ x - 2 $  will be factor of  $ p\left( x \right) = 3{x^2} + 6x - 24 $ , then  $ p\left( 2 \right) $  should be equal to  $ 0 $ .

Now,  $ p\left( 2 \right) = 3{\left( 2 \right)^2} + 6\left( 2 \right) - 24 $ 

 $  = 3\left( 4 \right) + 6\left( 2 \right) - 24 = 12 + 12 - 24 = 0 $ 

Therefore, by factor theorem  $ x - 2 $ , is factor of  $ 3{x^2} + 6x - 24 $ .

(ii)  $ 4{x^2} + x - 2 $

Let  $ p\left( x \right) = 4{x^2} + x - 2 $ 

If  $ x - 2 $  will be factor of  $ p\left( x \right) = 4{x^2} + x - 2 $ , then  $ p\left( 2 \right) $  should be equal to  $ 0 $ .

 $ p\left( 2 \right) = 4{\left( 2 \right)^2} + 2 - 2 $ 

 $  = 16 + 2 - 2 = 16 $ 

Since,  $ 16 \ne 0 $ 

Therefore,  $ x - 2 $  is not a factor of  $ 4{x^2} + x - 2 $ .

18. Show that  $ p - 1 $  is a factor of  $ {p^{10}} - 1 $  and also of  $ {p^{11}} - 1 $ .

Ans:   $ p - 1 $  is a factor  $ {p^{10}} - 1 $ , then  $ {\left( 1 \right)^{10}} - 1 $ should be equal to zero.

Now,  $ {\left( 1 \right)^{10}} - 1 = 1 - 1 = 0 $ 

Therefore,  $ p - 1 $  is a factor  $ {p^{10}} - 1 $ .

Again, if  $ p - 1 $  is a factor of  $ {p^{11}} - 1 $  , then  $ {\left( 1 \right)^{11}} - 1 $  should be equal to zero.

Now,  $ {\left( 1 \right)^{11}} - 1 = 1 - 1 = 0 $ 

Therefore,  $ p - 1 $ is a factor  $ {p^{11}} - 1 $ .

Hence,  $ p - 1 $  is a factor of  $ {p^{10}} - 1 $  and also of  $ {p^{11}} - 1 $ .

19. For what value of m is  $ {x^3} - 2m{x^2} + 16 $  divisible by  $ x + 2 $ ?

Ans: If  $ {x^3} - 2m{x^2} + 16 $  is divisible by  $ x + 2 $  , then  $ x + 2 $  is a factor of  $ {x^3} - 2m{x^2} + 16 $ .

Now, let  $ p\left( x \right) = {x^3} - 2m{x^2} + 16 $ 

As,  $ x + 2 = x - \left( { - 2} \right) $  is a factor of  $ {x^3} - 2m{x^2} + 16 $ .

So,  $ p\left( { - 2} \right) = 0 $ 

Now,  $ p\left( { - 2} \right) = {\left( { - 2} \right)^3} - 2m{\left( { - 2} \right)^2} + 16 $ 

 $  =  - 8 - 8m + 16 = 8 - 8m $ 

Now,  $ p\left( { - 2} \right) = 0 $ 

 $  \Rightarrow \,8 - 8m = 0 $ 

 $  \Rightarrow 8m = 8 $ 

 $  \Rightarrow m = 1 $ 

Hence, for  $ m = 1 $  ,  $ x + 2 $  is a factor of  $ {x^3} - 2m{x^2} + 16 $ so that  $ {x^3} - 2m{x^2} + 16 $  is completely divisible by  $ x + 2 $ .

20. If  $ x + 2a $  is a factor of  $ {x^5} - 4{a^2}{x^3} + 2x + 2a + 3 $ , find a.

Ans: Let  $ p\left( x \right) = {x^5} - 4{a^2}{x^3} + 2x + 2a + 3 $ 

If  $ x - \left( { - 2a} \right) $  is a factor of  $ p(x) $ , then  $ p\left( { - 2a} \right) = 0 $ .

 $ p\left( { - 2a} \right) = {\left( { - 2a} \right)^5} - 4{a^2}{\left( { - 2a} \right)^3} + 2\left( { - 2a} \right) + 2a + 3 $ 

 $  =  - 32{a^5} + 32{a^5} - 4a + 2a + 3 $ 

 $  =  - 2a + 3 $ 

Now,  $ p\left( { - 2a} \right) = 0 $ 

 $  \Rightarrow  - 2a + 3 = 0 $ 

 $  \Rightarrow a = \dfrac{3}{2} $ 

21. Find the value of m so that  $ 2x - 1 $  be a factor of  $ 8{x^4} + 4{x^3} - 16{x^2} + 10x + m $ .

Ans: Let  $ p\left( x \right) = 8{x^4} + 4{x^3} - 16{x^2} + 10x + m $ 

As,  $ \left( {2x - 1} \right) $  is a factor of  $ p\left( x \right) $ .

Then  $ p\left( {\dfrac{1}{2}} \right) = 0 $ (using factor theorem)

 $  \Rightarrow 8{\left( {\dfrac{1}{2}} \right)^4} + 4{\left( {\dfrac{1}{2}} \right)^3} - 16{\left( {\dfrac{1}{2}} \right)^2} + 10\left( {\dfrac{1}{2}} \right) + m = 0 $ 

 $  \Rightarrow 8\left( {\dfrac{1}{{16}}} \right) + 4\left( {\dfrac{1}{8}} \right) - 16\left( {\dfrac{1}{4}} \right) + 5 + m = 0 $ 

 $  \Rightarrow \dfrac{1}{2} + \dfrac{1}{2} - 4 + 5 + m = 0 $ 

 $  \Rightarrow 1 + 1 + m = 0 $ 

 $  \Rightarrow \,m =  - 2 $ 

22. If  $ x + 1 $  is a factor of  $ a{x^3} + {x^2} - 2x + 4a - 9 $ , find the value of a.

Ans: Let  $ p\left( x \right) = a{x^3} + {x^2} - 2x + 4a - 9 $ 

As  $ x + 1 $  is a factor of p(x)

 $ p\left( { - 1} \right) = 0 $  (By factor theorem)

 $  \Rightarrow \,a{\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - 2\left( { - 1} \right) + 4a - 9 = 0 $ 

 $  \Rightarrow \, - a + 1 + 2 + 4a - 9 = 0 $ 

 $  \Rightarrow \,3a - 6 = 0 $ 

 $  \Rightarrow \,3a = 6 $ 

 $  \Rightarrow \,a = 2 $ 

23. Factorize:

(i)  $ {x^2} + 9x + 18 $ 

Ans: In order to factorize  $ {x^2} + 9x + 18 $ , we have to split the middle term.

 $  \Rightarrow {x^2} + 9x + 18 $ 

 $  \Rightarrow {x^2} + \left( {6 + 3} \right)x + 18 $ 

 $  \Rightarrow {x^2} + 6x + 3x + 18 $ 

 $  \Rightarrow x\left( {x + 6} \right) + 3\left( {x + 6} \right) $ 

 $  \Rightarrow \left( {x + 3} \right)\left( {x + 6} \right) $ 

(ii)  $ 6{x^2} + 7x - 3 $ 

Ans: In order to factorize  $ 6{x^2} + 7x - 3 $ , we have to split the middle term.

 $  \Rightarrow 6{x^2} + 7x - 3 $ 

 $  \Rightarrow 6{x^2} + \left( {9 - 2} \right)x - 3 $ 

 $  \Rightarrow 6{x^2} + 9x - 2x - 3 $ 

 $  \Rightarrow 3x\left( {2x + 3} \right) - 1\left( {2x + 3} \right) $ 

 $  \Rightarrow \left( {3x - 1} \right)\left( {2x + 3} \right) $ 

(iii)  $ 2{x^2} - 7x - 15 $ 

Ans:  If we want to factorise  $ 2{x^2} - 7x - 15 $ , we have to split the middle term.

 $  \Rightarrow 2{x^2} - 7x - 15 $ 

 $  \Rightarrow 2{x^2} - \left( {10 - 3} \right)x - 15 $ 

 $  \Rightarrow 2{x^2} - 10x + 3x - 15 $ 

 $  \Rightarrow 2x\left( {x - 5} \right) + 3\left( {x - 5} \right) $ 

 $  \Rightarrow \left( {2x + 3} \right)\left( {x - 5} \right) $ 

(iv)  $ 84 - 2r - 2{r^2} $ 

Ans: If we want to factorise  $ 84 - 2r - 2{r^2} $ , we have to split the middle term.

 $  \Rightarrow  - \left( {2{r^2} + 2r - 84} \right) $ 

 $  \Rightarrow  - 2\left( {{r^2} + r - 42} \right) $ 

 $  \Rightarrow  - 2\left( {{r^2} + \left( {7 - 6} \right)r - 42} \right) $ 

 $  \Rightarrow  - 2\left( {{r^2} + 7r - 6r - 42} \right) $ 

 $  \Rightarrow  - 2\left( {r\left( {r + 7} \right) - 6\left( {r + 7} \right)} \right) $ 

 $  \Rightarrow 2\left( {6 - r} \right)\left( {r + 7} \right) $ 

24. Factorise:

(i)  $ 2{x^3} - 3{x^2} - 17x + 30 $ 

Ans: Let  $ f\left( x \right) = 2{x^3} - 3{x^2} - 17x + 30 $  be the given Polynomial. The factors of the constant term  $  + 30 $  are  $  \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 $  . The factor of coefficient of  $ {x^3} $  is  $ 2 $ .

Therefore, possible rational roots of f(x) are:

 $  \Rightarrow  \pm 1, \pm 3, \pm 5, \pm 15, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm \dfrac{5}{2}, \pm \dfrac{{15}}{2} $ 

We  $ f\left( 2 \right) = 2{\left( 2 \right)^3} - 3{\left( 2 \right)^2} - 17\left( 2 \right) + 30 $ 

 $  = 2\left( 8 \right) - 3\left( 4 \right) - 17\left( 2 \right) + 30 $ 

 $  = 16 - 12 - 34 + 30 = 0 $ 

And  $ f\left( { - 3} \right) = 2{\left( { - 3} \right)^3} - 3{\left( { - 3} \right)^2} - 17\left( { - 3} \right) + 30 $ 

 $  = 2\left( { - 27} \right) - 3\left( 9 \right) - 17\left( { - 3} \right) + 30 $ 

 $  =  - 54 - 27 + 51 + 30 = 0 $ 

Hence,  $ \left( {x - 2} \right) $   $ \left( {x + 3} \right) $ ,  will be the factors of  $ f\left( x \right) $ .

 $  \Rightarrow {x^2} + x - 6 $  is a factor of 

 $ f\left( x \right) $ .

Let's divide  $ f\left( x \right) = 2{x^3} - 3{x^2} - 17x + 30 $  by  $ {x^2} + x - 6 $  to get the other factors of  $ f\left( x \right) $ .

Factors of  $ f\left( x \right) $ .

By long division, we have 

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 $ 2{x^3} - 3{x^2} - 17x + 30 = \left( {{x^2} + x - 6} \right)\left( {2x - 5} \right) $ 

 $  \Rightarrow \,2{x^3} - 3{x^2} - 17x + 30 = \left( {x - 2} \right)\left( {x + 3} \right)\left( {2x - 5} \right) $ 

Hence,   $ \,2{x^3} - 3{x^2} - 17x + 30 = \left( {x - 2} \right)\left( {x + 3} \right)\left( {2x - 5} \right) $ 

(ii)  $ {x^3} - 6{x^2} + 11x - 6 $ 

Ans: Let  $ f\left( x \right) = {x^3} - 6{x^2} + 11x - 6 $ be the given Polynomial. The factors of the constant term  $  - 6 $  are  $  \pm 1, \pm 2, \pm 3\,\,and\,\, \pm 6 $ . 

We  $ f\left( 1 \right) = {\left( 1 \right)^3} - 6{\left( 1 \right)^2} + 11\left( 1 \right) - 6 = 0 $ 

And  $ f\left( 2 \right) = {\left( 2 \right)^3} - 6{\left( 2 \right)^2} + 11\left( 2 \right) - 6 = 8 - 24 + 22 - 6 = 0 $ 

Hence,  $ \left( {x - 1} \right) $   $ \left( {x - 2} \right) $ ,  will be factors   $ f\left( x \right) $ .

 $  \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) = {x^2} - 3x + 2 $  will be a factor of  $ f\left( x \right) $ .

Let us divide  $ f\left( x \right) = {x^3} - 6{x^2} + 11x - 6 $  by  $ {x^2} - 3x + 2 $  to get the other factors of  $ f\left( x \right) $ .

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 $ {x^3} - 6{x^2} + 11x - 6 = \left( {{x^2} - 3x + 2} \right)\left( {x - 3} \right) $ 

 $  \Rightarrow {x^3} - 6{x^2} + 11x - 6 = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) $ 

Hence,   $ {x^3} - 6{x^2} + 11x - 6 = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) $ 

(iii)  $ {x^3} + {x^2} - 4x + 4 $ 

Ans: Let  $ f\left( x \right) = {x^3} + {x^2} - 4x - 4 $ be the given Polynomial. The factors of the constant term  $  - 4 $  are  $  \pm 1, \pm 2, \pm 4\, $ . 

We  $ f\left( { - 1} \right) = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} - 4\left( { - 1} \right) - 4 = 0 $ 

And  $ f\left( 2 \right) = {\left( 2 \right)^3} + {\left( 2 \right)^2} - 4\left( 2 \right) - 4 = 8 + 4 - 8 - 4 = 0 $ 

Hence,  $ \left( {x + 1} \right) $  ,  $ \left( {x - 2} \right) $  are factors of  $ f\left( x \right) $ .

 $  \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right) = {x^2} - x - 2 $  will be a factor of  $ f\left( x \right) $ .

Let's now divide  $ f\left( x \right) = {x^3} + {x^2} - 4x - 4 $  by  $ {x^2} - x - 2 $  to get the other factors of  $ f\left( x \right) $ .

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 $ {x^3} + {x^2} - 4x - 4 = \left( {{x^2} - x - 2} \right)\left( {x + 2} \right) $ 

 $  \Rightarrow {x^3} + {x^2} - 4x - 4 = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right) $ 

Hence,   $ {x^3} + {x^2} - 4x - 4 = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right) $ 

(iv)  $ 3{x^3} - {x^2} - 3x + 1 $ 

Ans: Let  $ f\left( x \right) = 3{x^3} - {x^2} - 3x + 1 $ be the given Polynomial. The factors of a constant term  $  + 1 $  are  $  \pm 1.\, $ The factor of coefficient of  $ {x^3} $  is  $ 3 $ .Hence, possible rational roots of  $ f\left( x \right) $  are  $  \pm \dfrac{1}{3} $ . 

 $ f\left( 1 \right) = 3{\left( 1 \right)^3} - {\left( 1 \right)^2} - 3\left( 1 \right) + 1 = 0 $ 

And  $ f\left( { - 1} \right) = 3{\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} - 3\left( { - 1} \right) + 1 =  - 3 - 1 + 3 + 1 = 0 $ 

Hence,  $ \left( {x - 1} \right) $   $ \left( {x + 1} \right) $ ,  will be factors of  $ f\left( x \right) $ .

 $  \Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) = {x^2} - 1 $  will be factor of  $ f\left( x \right) $ .

Let's now divide  $ f\left( x \right) = 3{x^3} - {x^2} - 3x + 1 $  by  $ {x^2} - 1 $  to get the other factors of  $ f\left( x \right) $ .

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 $ 3{x^3} - {x^2} - 3x + 1 = \left( {{x^2} - 1} \right)\left( {3x - 1} \right) $ 

 $  \Rightarrow 3{x^3} - {x^2} - 3x + 1 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {3x - 1} \right) $ 

Hence,   $ 3{x^3} - {x^2} - 3x + 1 = \left( {x - 1} \right)\left( {x + 1} \right)\left( {3x - 1} \right) $ 

25. Using suitable identify, evaluate the following:

(i)  $ {103^3} $ 

Ans: $ {103^3} = {\left( {100 + 3} \right)^3} $ 

Now using identify  $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ , we have

 $ {\left( {100 + 3} \right)^3} = {\left( {100} \right)^3} + {\left( 3 \right)^3} + 3\left( {100} \right)\left( 3 \right)\left( {100 + 3} \right) $ 

 $  = 1000000 + 27 + 900\left( {100 + 3} \right) $ 

 $  = 1000000 + 27 + 90000 + 2700 $ 

 $  = 1092727 $ 

(ii)  $ 101 \times 102 $ 

Ans: $ 101 \times 102 $ 

 $  = \left( {100 + 1} \right)\left( {100 + 2} \right) $  (using  $ \left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab $ )

 $  = {\left( {100} \right)^2} + \left( {1 + 2} \right)100 + \left( 1 \right)\left( 2 \right) $ 

 $  = 10000 + \left( 3 \right)100 + 2 = 10000 + 300 + 2 = 10302 $ 

(iii)  $ {999^2} $ 

Ans: $ {999^2} $ 

Ans:  $ {\left( {1000 - 1} \right)^2} $ 

 $  = {\left( {1000} \right)^2} - 2\left( {1000} \right) \times 1 + {1^2} $ 

 $  = 1000000 - 2000 + 1 = 998001 $ 

26. Factorise the following:

(i)  $ 4{x^2} + 20x + 25 $ 

Ans: We have,

 $ 4{x^2} + 20x + 25 = {\left( {2x} \right)^2} + 2\left( {2x} \right)5 + {\left( 5 \right)^2} $ 

 $  = {\left( {2x + 5} \right)^2} $   $ \left[ {{a^2} + 2ab + {b^2} = {{\left( {a + b} \right)}^2}} \right] $ 

 $  = \left( {2x + 5} \right)\left( {2x + 5} \right) $ 

(ii)  $ 9{y^2} - 66yz + 121{z^2} $ 

 $ 9{y^2} - 66yz + 121{z^2} = {\left( {3y} \right)^2} - 2\left( {3y} \right)\left( {11z} \right) + {\left( { - 11z} \right)^2} $ 

 $  = {\left( {3y - 11z} \right)^2} $   $ \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] $ 

 $  = \left( {3y - 11} \right)\left( {3y - 11} \right) $ 

(iii)  $ {\left( {2x + \dfrac{1}{3}} \right)^2} - {\left( {x - \dfrac{1}{2}} \right)^2} $ 

Ans: $ {\left( {2x + \dfrac{1}{3}} \right)^2} - {\left( {x - \dfrac{1}{2}} \right)^2} $ 

Now, using identity  $ {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right) $ 

 $  = \left[ {\left( {2x + \dfrac{1}{3}} \right) + \left( {x - \dfrac{1}{2}} \right)} \right]\left[ {\left( {2x + \dfrac{1}{3}} \right) - \left( {x - \dfrac{1}{2}} \right)} \right] $ 

 $  = \left( {2x + \dfrac{1}{3} + x - \dfrac{1}{2}} \right)\left( {2x + \dfrac{1}{3} - x + \dfrac{1}{2}} \right) = \left( {3x - \dfrac{1}{6}} \right)\left( {x + \dfrac{5}{6}} \right) $ 

27. Factorise the following:

(i)  $ 9{x^2} - 12x + 3 $ 

Ans: we have,

 $ 9{x^2} - 12x + 3 = 9{x^2} - 9x - 3x + 3 $ 

 $  = 9x\left( {x - 1} \right) - 3\left( {x - 1} \right) $ 

 $  = \left( {9x - 3} \right)\left( {x - 1} \right) $ 

 $  = 3\left( {3x - 1} \right)\left( {x - 1} \right) $ 

(ii)  $ 9{x^2} - 12x + 4 $ 

 $ 9{x^2} - 12x + 4 = {\left( {3x} \right)^2} - 2\left( {3x} \right)\left( 2 \right) + {\left( 2 \right)^2} $ 

 $  = {\left( {3x - 2} \right)^2}\left[ {{a^2} - 2ab + {b^2} = {{\left( {a - b} \right)}^2}} \right] $ 

 $  = \left( {3x - 2} \right)\left( {3x - 2} \right) $ 

28. Expand the following:

(i)  $ {\left( {4a - b + 2c} \right)^2} $ 

 $ {\left( {4a - b + 2c} \right)^2} = {\left( {4a} \right)^2} + {\left( { - b} \right)^2} + {\left( {2c} \right)^2} + 2\left( {4a} \right)\left( { - b} \right) + 2\left( { - b} \right)\left( {2c} \right) + 2\left( {2c} \right)\left( {4a} \right) $ 

Using  $ \left[ {{a^2} + b + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $ 

 $  = 16{a^2} + {b^2} + 4{c^2} - 8ab - 4bc + 16ca $ 

(ii)  $ {\left( {3a - 5b - c} \right)^2} $ 

 $ {\left( {3a - 5b - c} \right)^2} = {\left( {3a} \right)^2} + {\left( { - 5b} \right)^2} + {\left( { - c} \right)^2} + 2\left( {3a} \right)\left( { - 5b} \right) + 2\left( { - 5b} \right)\left( { - c} \right) + 2\left( { - c} \right)\left( {3a} \right) $ 

Using  $ \left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $ 

 $  = 9{a^2} + 25{b^2} + {c^2} - 30ab + 10bc - 6ca $ 

(iii)  $ {\left( { - x + 2y - 3z} \right)^2} $ 

 $ {\left( { - x + 2y - 3z} \right)^2} = {\left( { - x} \right)^2} + {\left( {2y} \right)^2} + {\left( { - 3z} \right)^2} + 2\left( { - x} \right)\left( {2y} \right) + 2\left( {2y} \right)\left( { - 3z} \right) + 2\left( { - 3z} \right)\left( { - x} \right) $ 

 $  = {x^2} + 4{y^2} + 9{z^2} - 4xy - 12yz + 6xz $ 

29. Factorise the following:

(i)  $ 9{x^2} + 4{y^2} + 16{z^2} + 12xy - 16yz - 24xz $ 

 $  \Rightarrow {\left( {3x} \right)^2} + {\left( {2y} \right)^2} + {\left( { - 4z} \right)^2} + 2\left( {3x} \right)\left( {2y} \right) + 2\left( {2y} \right)\left( { - 4z} \right) + 2\left( { - 4z} \right)\left( {3x} \right) $ 

 $  \Rightarrow {\left\{ {3x + 2y + \left( { - 4z} \right)} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $ 

 $  \Rightarrow {\left( {3x + 2y - 4z} \right)^2} = \left( {3x + 2y - 4z} \right)\left( {3x + 2y - 4z} \right) $ 

(ii)  $ 25{x^2} + 16{y^2} + 4{z^2} - 40xy + 16yz - 20xz $ 

 $  \Rightarrow {\left( { - 5x} \right)^2} + {\left( {4y} \right)^2} + {\left( {2z} \right)^2} + 2\left( { - 5x} \right)\left( {4y} \right) + 2\left( {4y} \right)\left( {2z} \right) + 2\left( {2z} \right)\left( { - 5x} \right) $ 

 $  \Rightarrow {\left\{ { - 5x + 4y + 2z} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $ 

(iii)  $ 16{x^2} + 4{y^2} + 9{z^2} - 16xy - 12yz + 24xz $ 

 $  \Rightarrow {\left( {4x} \right)^2} + {\left( { - 2y} \right)^2} + {\left( {3z} \right)^2} + 2\left( {4x} \right)\left( { - 2y} \right) + 2\left( { - 2y} \right)\left( {3z} \right) + 2\left( {3z} \right)\left( {4x} \right) $ 

 $  \Rightarrow {\left\{ {4x - 2y + 3z} \right\}^2}\,\left[ {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {{\left( {a + b + c} \right)}^2}} \right] $ 

 $  \Rightarrow {\left( {4x - 2y + 3z} \right)^2} = \left( {4x - 2y + 3z} \right)\left( {4x - 2y + 3z} \right) $ 

30. If  $ a + b + c = 9 $  and  $ ab + bc + ca = 26 $ , find  $ {a^2} + {b^2} + {c^2} $ .

Ans: We know that 

 $ {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca $ 

 $  \Rightarrow {\left( {a + b + c} \right)^2} = \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {ab + bc + ca} \right) $ 

Now, on putting the values, we get

 $  \Rightarrow {\left( 9 \right)^2} = \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {26} \right) $ 

 $  \Rightarrow {a^2} + {b^2} + {c^2} = 81 - 52 = 29 $ 

31. Expand the following:

(i)  $ {\left( {3a - 2b} \right)^3} $ 

 $ {\left( {3a - 2b} \right)^3} = {\left( {3a} \right)^3} - {\left( {2b} \right)^3} - 3\left( {3a} \right)\left( {2b} \right)\left( {3a - 2b} \right) $ 

Now, using identity  $ \left[ {{{\left( {a - b} \right)}^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)} \right] $ 

 $  = 27{a^3} - 8{b^3} - 54{a^2}b + 36a{b^2} $ 

(ii)  $ {\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right)^3} $ 

Ans: Here, using the identity  $ \left[ {{{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] $ 

 $ {\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right)^3} = {\left( {\dfrac{1}{x}} \right)^3} + {\left( {\dfrac{y}{3}} \right)^3} + 3 \times \dfrac{1}{x} \times \dfrac{y}{3}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $ 

 $  \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{x}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $ 

 $  \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{{{x^2}}} + \dfrac{{{y^2}}}{{3x}} $ 

(iii)  $ {\left( {4 - \dfrac{1}{{3x}}} \right)^3} $ 

 $ {\left( {4 - \dfrac{1}{{3x}}} \right)^3} = {\left( 4 \right)^3} - {\left( {\dfrac{1}{{3x}}} \right)^3} - 3\left( 4 \right)\left( {\dfrac{1}{{3x}}} \right)\left( {4 - \dfrac{1}{{3x}}} \right) $ 

 $  = 64 - \dfrac{1}{{27{x^3}}} - \dfrac{4}{x}\left( {4 - \dfrac{1}{{3x}}} \right) $ 

 $  = 64 - \dfrac{1}{{27{x^3}}} - \dfrac{{16}}{x} + \dfrac{4}{{3{x^2}}} $ 

32.Factorise the following:

(i)  $ 1 - 64{a^3} - 12a + 48{a^2} $ 

 $ 1 - 64{a^3} - 12a + 48{a^2} = {\left( 1 \right)^3} - {\left( {4a} \right)^3} - 3\left( 1 \right)\left( {4a} \right)\left( {1 - 4a} \right) $ 

 $  = {\left( {1 - 4a} \right)^3}\left[ {{a^3} - {b^3} - 3ab\left( {a - b} \right) = {{\left( {a - b} \right)}^3}} \right] $ 

 $  = \left( {1 - 4a} \right)\left( {1 - 4a} \right)\left( {1 - 4a} \right) $ 

(ii)  $ 8{p^3} + \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p + \dfrac{1}{{125}} $ 

$ 8{p^3} + \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p + \dfrac{1}{{125}} $ 

 $  = {\left( {2p} \right)^3} + 3 \times {\left( {2p} \right)^2} \times \dfrac{1}{5} + 3 \times \left( {2p} \right) \times {\left( {\dfrac{1}{5}} \right)^2} + {\left( {\dfrac{1}{5}} \right)^3} $ 

 $  = {\left( {2p} \right)^3} + {\left( {\dfrac{1}{5}} \right)^3} + 3 \times \left( {2p} \right) \times \dfrac{1}{5}\left[ {2p + \dfrac{1}{5}} \right] $ 

 $  = {\left( {2p + \dfrac{1}{5}} \right)^3} = \left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right) $ 

33. Find the following products:

(i)  $ \left( {\dfrac{x}{2} + 2y} \right)\left( {\dfrac{{{x^2}}}{4} - xy + 4{y^2}} \right) $ 

 $ \left( {\dfrac{x}{2} + 2y} \right)\left( {\dfrac{{{x^2}}}{4} - xy + 4{y^2}} \right) = \left( {\dfrac{x}{y} + 2y} \right)\left( {{{\left( {\dfrac{x}{2}} \right)}^2} - \left( {\dfrac{x}{2}} \right)\left( {2y} \right) + {{\left( {2y} \right)}^2}} \right) $ 

 $  = {\left( {\dfrac{x}{2}} \right)^3} + {\left( {2y} \right)^3}\left[ {\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3}} \right] $ 

 $  = \dfrac{{{x^3}}}{8} + 8{y^3} $ 

(ii)  $ \left( {{x^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right) $ 

 $ \left( {{{\left( x \right)}^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right) = \left( {{x^2} - 1} \right)\left( {{{\left( {{x^2}} \right)}^2} + \left( {{x^2}} \right)\left( 1 \right) + {{\left( 1 \right)}^2}} \right) $ 

 $  = {\left( {{x^2}} \right)^3} - {\left( 1 \right)^3}\left[ {\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3}} \right] $ 

 $  = {x^6} - 1 $ 

34. Factorise:

(i)  $ 1 + 64{x^3} $ 

 $ 1 + 64{x^3} = {\left( 1 \right)^3} + {\left( {4x} \right)^3} $ 

 $  = \left( {1 + 4x} \right)\left( {{{\left( 1 \right)}^2} - \left( 1 \right)\left( {4x} \right) + {{\left( {4x} \right)}^2}} \right) $ 

 $  = \left( {1 + 4x} \right)\left( {1 - 4x + 16{x^2}} \right) $ 

(ii)  $ {a^3} - 2\sqrt {2{b^3}}  $ 

 $ {a^3} - 2\sqrt {2{b^3}}  = {\left( a \right)^3} - {\left( {\sqrt 2 b} \right)^3} $ 

 $  = \left( {a - \sqrt 2 b} \right)\left( {{{\left( a \right)}^2} + \left( a \right)\left( {\sqrt 2 b} \right) + {{\left( {\sqrt 2 b} \right)}^2}} \right) $ 

 $  = \left( {a - \sqrt 2 b} \right)\left( {{a^2} + \sqrt 2 ab + 2{b^2}} \right) $ 

35. Find the following product:

 $ \left( {2x - y + 3z} \right)\left( {4{x^2} + {y^2} + 9{z^2} + 2xy + 3yz - 6xz} \right) $ 

 $  = \left\{ {2x + \left( { - y} \right) + 3z} \right\}\left\{ {{{\left( {2x} \right)}^2} + {{\left( { - y} \right)}^2} + {{\left( {3z} \right)}^2} - 2x\left( { - y} \right) - \left( { - y} \right)\left( {3z} \right) - \left( { - y} \right)\left( {3z} \right) - \left( {3z} \right)\left( {2x} \right)} \right\} $ 

 $  = {\left( {2x} \right)^3} + {\left( { - y} \right)^3} + {\left( {3z} \right)^3} - 3\left( {2x} \right)\left( { - y} \right)\left( {3z} \right) $  

using identity  $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  = 8{x^3} - {y^3} + 27{z^2} + 18xyz $ 

36.Factorise:

(i)  $ {a^3} - 8{b^3} - 64{c^3} - 24abc $ 

 $ {a^3} - 8{b^3} - 64{c^3} - 24abc $ 

 $  = \left\{ {{{\left( a \right)}^3} + {{\left( { - 2b} \right)}^3} + {{\left( { - 4c} \right)}^3} - 3\left( a \right)\left( { - 2b} \right)\left( { - 4c} \right)} \right\} $ 

 $  = \left\{ {a + \left( { - 2b} \right) + \left( { - 4c} \right)} \right\}\left\{ {{a^2} + {{\left( { - 2b} \right)}^2} + {{\left( { - 4c} \right)}^2} - a\left( { - 2b} \right) - \left( { - 2b} \right)\left( { - 4c} \right) - \left( { - 4c} \right)a} \right\} $ 

Using identity  $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  = \left( {a - 2b - 4c} \right)\left( {{a^2} + 4{b^2} + 16{c^2} + 2ab - 8bc + 4ca} \right) $ 

(ii)  $ 2\sqrt 2 {a^3} + 8{b^3} - 27{c^3} + 18\sqrt 2 abc $ 

 $ 2\sqrt 2 {a^3} + 8{b^3} - 27{c^3} + 18\sqrt 2 abc $ 

 $  = \left\{ {{{\left( {\sqrt 2 a} \right)}^3} + {{\left( {2b} \right)}^3} + {{\left( { - 3c} \right)}^3} - 3\left( {\sqrt 2 a} \right)\left( {2b} \right)\left( { - 3c} \right)} \right\} $ 

 $  = \left\{ {\sqrt 2 a + 2b + \left( { - 3c} \right)} \right\}\left\{ {{{\left( {\sqrt 2 a} \right)}^2} + {{\left( {2b} \right)}^2} + {{\left( { - 3c} \right)}^2} - \left( {\sqrt 2 a} \right)\left( {2b} \right) - \left( {2b} \right)\left( { - 3c} \right) - \left( { - 3c} \right)\left( {\sqrt 2 a} \right)} \right\} $ 

 $  = \left( {\sqrt 2 a + 2b - 3c} \right)\left( {2{a^2} + 4{b^2} + 9{c^2} - 2\sqrt 2 ab + 6bc + 3\sqrt 2 ca} \right) $ 

37. Without actually calculating the cubes, find the value of:

(i)  $ {\left( {\dfrac{1}{2}} \right)^3} + {\left( {\dfrac{1}{3}} \right)^3} - {\left( {\dfrac{5}{6}} \right)^3} $ 

Ans:  Let  $ a = \dfrac{1}{2}\,,\,b = \dfrac{1}{3}\,,\,c =  - \dfrac{5}{6} $ 

 $ a + b + c = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{5}{6} $ 

 $  = \dfrac{{3 + 2 - 5}}{6} = \dfrac{0}{6} = 0 $ 

 $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  \Rightarrow 3 \times \dfrac{1}{2} \times \dfrac{1}{3} \times  - \dfrac{5}{6} $ 

 $  \Rightarrow  - \dfrac{5}{{12}} $ 

(ii)  $ {\left( {0.2} \right)^3} - {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} $ 

 $ {\left( {0.2} \right)^3} - {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} = {\left( {0.2} \right)^3} + {\left( { - 0.3} \right)^3} + {\left( {0.1} \right)^3} $ 

 $ a = 0.2\,,\,b =  - 0.3\,,\,c = 0.1 $ . Then,

 $ a + b + c = 0.2 - 0.3 - 0.1 = 0 $ .

 $  \Rightarrow 3 \times 0.2 \times \left( { - 0.3} \right) \times \left( {0.1} \right) $ 

 $  \Rightarrow  - 0.018 $ 

Hence,  $ {\left( {0.2} \right)^3} - {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} =  - 0.018 $ 

38. Without finding the cubes, factorise

 $ {\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3} $ 

 $ a = x - 2y\,,\,b = 2y - 3z\,,\,c = 3z - x $ . Then,

 $ a + b + c = x - 2y + 2y - 3z + 3z - x = 0 $ .

 $  \Rightarrow 3 \times \left( {x - 2y} \right) \times \left( {2y - 3z} \right) \times \left( {3z - x} \right) $ 

 $  \Rightarrow 3\left( {x - 2y} \right)\left( {2y - 3z} \right)\left( {3z - x} \right) $ 

39. Find the value of

(i)  $ {x^3} + {y^3} - 12xy + 64 $ , when  $ x + y =  - 4 $ 

Ans: $ {x^3} + {y^3} - 12xy + 64 = {x^3} + {y^3} + {4^3} - 3xy\left( 4 \right) $ 

 $  = \left( {x + y + 4} \right)\left( {{x^2} + {y^2} + {4^2} - xy - 4y - 4x} \right) $ 

 $ \left[ {x + y =  - 4} \right] $ 

 $  = \left( 0 \right)\left( {{x^2} + {y^2} + {4^2} - xy - 4y - 4x} \right) = 0 $ 

(ii)  $ {x^3} - 8{y^3} - 36xy - 216 $ , when  $ x = 2y + 6 $ 

Ans: $ {x^3} - 8{y^3} - 36xy - 216 = {x^3} + {\left( { - 2y} \right)^3} + {\left( { - 6} \right)^3} - 3x\left( { - 2y} \right)\left( { - 6} \right) $ 

 $  = \left( {x - 2y - 6} \right)\left( {{x^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - 6} \right)}^2} - x\left( { - 2y} \right) - \left( { - 2y} \right)\left( { - 6} \right) - x\left( { - 6} \right)} \right) $ 

 $  = \left( {x - \left( {2y + 6} \right)} \right)\left( {{x^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - 6} \right)}^2} - x\left( { - 2y} \right) - \left( { - 2y} \right)\left( { - 6} \right) - x\left( { - 6} \right)} \right) $ 

 $ \left[ {x = 2y + 6} \right] $ 

 $  = \left( 0 \right)\left( {{x^2} + 4{y^2} + 36 + 2xy - 12y + 6x} \right) = 0 $ 

40. Give possible experiments for the length and breadth of the rectangle whose area is given by  $ 4{a^2} + 4a - 3 $ 

Ans: Area  $ 4{a^2} + 4a - 3 $ .

Here we will use a method of splitting the middle term.

Here we have to split the middle term as  $ 4a = 6a - 2a $ 

  $  \Rightarrow 4{a^2} + \left( {6a - 2a} \right) - 3 $ 

 $  \Rightarrow 4{a^2} + 6a - 2a - 3 $ 

 $  \Rightarrow 2a\left( {2a + 3} \right) - 1\left( {2a + 3} \right) $ 

 $  \Rightarrow \left( {2a - 1} \right)\left( {2a + 3} \right) $ 

We know that, area of rectangle \[ = 4{a^2} + 4a - 3\]

Here we also know that, area of a rectangle  $  = \,length\, \times \,breadth $  and  $ 4{a^2} + 4a - 3 = \left( {2a - 1} \right)\left( {2a + 3} \right) $ .

Therefore, its possible length and breadth  $  = \,\left( {2a - 1} \right) $  and   $ \left( {2a + 3} \right) $  or, we can say that  $ length = \left( {2a + 3} \right) $  and  $ breadth\, = \,\left( {2a - 1} \right) $ .

Long Answer Questions

Sample Question 1.  If \[x{\text{ }} + {\text{ }}y{\text{ }} = {\text{ }}12\] and \[xy{\text{ }} = {\text{ }}27\], find the value of \[{x^3}{\text{ }} + {\text{ }}{y^3}\].

Ans: Here we will use an identity, \[{\text{ }}{\left( {x{\text{ }} + {\text{ }}y} \right)^3}{\text{ =  }}{x^3}{\text{ }} + {\text{ }}{y^3}{\text{  +  }}3xy{\text{ }}\left( {x{\text{ }} + {\text{ }}y} \right)\]

\[{x^3}{\text{ }} + {\text{ }}{y^3}{\text{ }} = {\text{ }}{\left( {x{\text{ }} + {\text{ }}y} \right)^3}{\text{ }}-{\text{ }}3xy{\text{ }}\left( {x{\text{ }} + {\text{ }}y} \right)\] 

\[{\text{ =  }}{12^3}{\text{ }}-{\text{ }}3{\text{ }} \times {\text{ }}27{\text{ }} \times {\text{ }}12\]

\[ = {\text{ }}12{\text{ }}\left[ {{{12}^2}{\text{ }}-{\text{ }}3{\text{ }} \times {\text{ }}27} \right]\]

\[ = {\text{ }}12{\text{ }} \times {\text{ }}63\]

\[ = {\text{ }}756\]

Therefore, the value of \[{x^3}{\text{ }} + {\text{ }}{y^3} = 756\] .

EXERCISE 2.4   

1. If the Polynomials  $ a{z^3} + 4{z^2} + 3z - 4 $  and  $ {z^3} - 4z + a $  leave the same remainder when divided  $ z - 3 $ , by Find the value of a.

 $ p\left( z \right) = a{z^3} + 4{z^2} + 3z - 4 $ 

And  $ q\left( z \right) = {z^3} - 4z + a $ 

As it is given that both the Polynomials leave the same remainder when divided by  $ z - 3 $ .

 $ p\left( 3 \right) = q\left( 3 \right) $ 

\[ \Rightarrow a{\left( 3 \right)^3} + 4{\left( 3 \right)^2} + 3\left( 3 \right) - 4 = \,{\left( 3 \right)^3} - 4\left( 3 \right) + a\]

\[ \Rightarrow 27a + 4 \times 9 + 9 - 4 = \,27 - 12 + a\]

\[ \Rightarrow 27a + 36 + 5 = \,15 + a\]

 $  \Rightarrow 27a - a = 15 - 41 $ 

 $  \Rightarrow 26a =  - 26 $ 

 $  \Rightarrow a =  - 1 $ 

Hence, the required value of a is  $  - 1 $ .

2. The Polynomial  $ p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + 3a - 7 $ when divided by  $ x + 1 $  leave remainder  $ 19 $ . Also, find the remainder when  $ p\left( x \right) $  is divided by  $ x + 2 $ .

Ans: We know that when  $ p\left( x \right) $  is divided by  $ x + b $ , then the remainder  $  = \,p\left( { - b} \right) $ 

Now,  $ p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - ax + 3a - 7 $ is divided by  $ x + 1 $ , then the remainder  $  = \,p\left( { - 1} \right) $ .

Also, we know that the remainder is  $ 19 $ .

 $ \,p\left( { - 1} \right) = 19 $ 

 $ {\left( { - 1} \right)^4} - 2{\left( { - 1} \right)^3} + 3{\left( { - 1} \right)^2} - a\left( { - 1} \right) + 3a - 7 = 19 $ 

 $  \Rightarrow 1 + 2 + 3 + a + 3a - 7 = 19 $ 

 $  \Rightarrow 4a - 1 = 19 $ 

 $  \Rightarrow 4a = 20 $ 

 $  \Rightarrow a = 5 $ 

Now,  $ p\left( x \right) = {x^4} - 2{x^3} + 3{x^2} - 5x + 3\left( 5 \right) - 7 = {x^4} - 2{x^3} + 3{x^2} - 5x + 8 $ 

Again, when p(x) is divided by  $ x + 2 $ , then

Remainder  $  = p\left( { - 2} \right) = {\left( { - 2} \right)^4} - 2{\left( { - 2} \right)^3} + 3{\left( { - 2} \right)^2} - 5\left( { - 2} \right) + 8 $ 

 $  = 16 + 16 + 12 + 10 + 8 $ 

 $  = 62 $ 

3. If both \[\left( {x - 2} \right)\] and  $ \left( {x - \dfrac{1}{2}} \right) $  are factors of  $ p{x^2} + 5x + r $ , Show that  $ p = r $ .

Ans: Let  $ q\left( x \right) = p{x^2} + 5x + r $ 

As  $ \left( {x - 2} \right) $  is a factor of  $ q\left( x \right) $ 

Then,  $ q\left( 2 \right) = 0 $ 

 $  \Rightarrow p{\left( 2 \right)^2} + 5\left( 2 \right) + r = 0 $ 

 $  \Rightarrow 4p + 10 + r = 0 $ 

 $  \Rightarrow  - 4p - r = 10............\left( 1 \right) $ 

Again,  $ \left( {x - \dfrac{1}{2}} \right) $  is a factor of  $ q\left( x \right) $ .

Then,  $ q\left( {\dfrac{1}{2}} \right) = 0 $ 

 $  \Rightarrow p{\left( {\dfrac{1}{2}} \right)^2} + 5\left( {\dfrac{1}{2}} \right) + r = 0 $ 

 $  \Rightarrow \dfrac{p}{4} + \dfrac{5}{2} + r = 0 $ 

 $  \Rightarrow p + 10 + 4r = 0 $ 

Now, using equation  $ 1 $ 

 $  \Rightarrow p - 4p - r + 4r = 0 $ 

 $  \Rightarrow  - 3p + 3r = 0 $ 

 $  \Rightarrow p = r $ 

Hence, proved.

4. Without actual division, prove that  $ 2{x^4} - 5{x^3} + 2{x^2} - x + 2 $  is divisible by  $ {x^2} - 3x + 2 $ .

 $ {x^2} - 3x + 2 $ 

We can also write it as

 $  \Rightarrow {x^2} - 3x + 2 = {x^2} - 2x - x + 2 $ 

 $  \Rightarrow x\left( {x - 2} \right) - 1\left( {x - 2} \right) = \left( {x - 1} \right)\left( {x - 2} \right) $ 

Now, we have

 $ p\left( x \right) = 2{x^4} - 5{x^3} + 2{x^2} - x + 2 $ 

Put  $ x = 1 $ 

 $ p\left( 1 \right) = 2{\left( 1 \right)^4} - 5{\left( 1 \right)^3} + 2{\left( 1 \right)^2} - \left( 1 \right) + 2 $ 

 $ p\left( 1 \right) = 2 - 5 + 2 + 1 = 0 $ 

 $ p\left( 1 \right) = 0 $ 

Therefore,  $ x - 1 $  divides p(x).

Put  $ x = 2 $ 

 $ p\left( 2 \right) = 2{\left( 2 \right)^4} - 5{\left( 2 \right)^3} + 2{\left( 2 \right)^2} - \left( 2 \right) + 2 $ 

 $ p\left( 2 \right) = 32 - 40 + 8 - 2 + 2 = 0 $ 

 $ p\left( 2 \right) = 0 $ 

Therefore,  $ x - 2 $  divides p(x).

Hence, $ {x^2} - 3x + 2 $ divides $ 2{x^4} - 5{x^3} + 2{x^2} - x + 2 $ .

5. Simplify  $ {\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} $ .

 $  = {\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} $ 

Now, using  $ {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) $ 

 $  = \left\{ {\left( {2x - 5y} \right) - \left( {2x + 5y} \right)} \right\}\left\{ {{{\left( {2x - 5y} \right)}^2} + \left( {2x - 5y} \right)\left( {2x + 5y} \right) + {{\left( {2x + 5} \right)}^2}} \right\} $ 

 $  = \left( { - 10y} \right)\left( {2{x^2} + 25{y^2}} \right) $ 

 $  =  - 120{x^2}y - 250{y^3} $ 

6. Multiply  $ {x^2} + 4{y^2} + {z^2} + 2xy + xz - 2yz $  by  $ \left( { - z + x - 2y} \right) $ .

 $ \left( { - z + x - 2y} \right)\left( {{x^2} + 4{y^2} + {z^2} + 2xy + xz - 2yz} \right) $ 

 $  = \left\{ {x + \left( { - 2y} \right) + \left( { - z} \right)} \right\}\left\{ {{{\left( x \right)}^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - z} \right)}^2} - \left( x \right)\left( { - 2y} \right) - \left( x \right)\left( { - z} \right) - \left( -2y \right)\left( { - z} \right)} \right\} $ 

 $  = {x^3} + {\left( { - 2y} \right)^3} + {\left( { - z} \right)^3} - 3\left( x \right)\left( { - 2y} \right)\left( { - z} \right) $ 

 $  = {x^3} - 8{y^3} - {z^3} - 6xyz $ 

7. If a, b, c are all non-zero and \[a + b + c = 0\], prove that  $ \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3 $ .

Ans: Here, a, b, c are all non-zero and \[a + b + c = 0\].

 $ {a^3} + {b^3} + {c^3} = 3abc $ 

Now, we will divide the complete equation by abc.

 $ \dfrac{{{a^3}}}{{abc}} + \dfrac{{{b^3}}}{{abc}} + \dfrac{{{c^3}}}{{abc}} = \dfrac{{3abc}}{{abc}} $ 

 $ \dfrac{{{a^2}}}{{bc}} + \dfrac{{{b^2}}}{{ca}} + \dfrac{{{c^2}}}{{ab}} = 3 $ .

8. If  $ a + b + c = 5 $  and  $ ab + bc + ca = 10 $ , then prove that  $ {a^3} + {b^3} + {c^3} - 3abc =  - 25 $ .

Ans: We know that, 

 $  = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ab + bc + ca} \right)} \right) $ 

It is given that  $ a + b + c = 5 $  and  $ ab + bc + ca = 10 $ .

 $  = 5\left( {{a^2} + {b^2} + {c^2} - 10} \right) $ 

Now, we have 

 $ a + b + c = 5 $ 

Squaring both sides, we get

 $ {\left( {a + b + c} \right)^2} = {\left( 5 \right)^2} $ 

 $ {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) = 25 $ 

 $ {a^2} + {b^2} + {c^2} + 2\left( {10} \right) = 25 $ 

 $ {a^2} + {b^2} + {c^2} = 25 - 20 = 5 $ 

Now, put the above value in equation  $  = 5\left( {{a^2} + {b^2} + {c^2} - 10} \right) $ 

 $  = 5\left( {5 - 10} \right) $ 

 $  = 5 \times  - 5 =  - 25 $ 

9. Prove that  $ {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) $ .

Ans:   $ {\left( {a + b + c} \right)^3} = {\left[ {a + \left( {b + c} \right)} \right]^3} $ 

 $  = {a^3} + 3{a^2}\left( {b + c} \right) + 3a{\left( {b + c} \right)^2} + {\left( {b + c} \right)^3} $ 

 $  = {a^3} + 3{a^2}b + 3{a^2}c + 3a\left( {{b^2} + 2bc + {c^2}} \right) + \left( {{b^3} + 3{b^2}c + 3b{c^2} + {c^3}} \right) $ 

 $  = {a^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + {b^3} + 3{b^2}c + 3b{c^2} + {c^3} $ 

 $  = {a^3} + {b^3} + {c^3} + 3{a^2}b + 3{a^2}c + 3a{b^2} + 6abc + 3a{c^2} + 3{b^2}c + 3b{c^2} $ 

 $  = {a^3} + {b^3} + {c^3} + 3abc + 3{a^2}b + 3a{c^2} + 3{a^2}c + 3{b^2}c + 3{b^2}a + 3b{c^2} + 3abc $ 

 $  = {a^3} + {b^3} + {c^3} + 3ab\left( {a + c} \right) + 3ac\left( {a + c} \right) + 3{b^2}\left( {a + c} \right) + 3bc\left( {a + c} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3ab + 3ac + 3{b^2} + 3bc} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {3a\left( {b + c} \right) + 3b\left( {b + c} \right)} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + \left( {a + c} \right)\left( {b + c} \right)\left( {3a + 3b} \right) $ 

 $  = {a^3} + {b^3} + {c^3} + 3\left( {a + c} \right)\left( {b + c} \right)\left( {a + b} \right) $ 

Now, on transposing we get

 $ {\left( {a + b + c} \right)^3} - {a^3} - {b^3} - {c^3} = 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) $ 

Introduction to NCERT Exemplar

NCERT Exemplar Class 9 Maths Solutions Chapter 2 'Polynomials' will give you data in regards to polynomials and their applications. The part of NCERT Exemplar Class 9 Maths incorporates various topics like entire numbers, integers, rational numbers, kinds of polynomials, zeros of polynomials, and so forth This part contains 4 activities and the evaluations of the around 60 model problems.

Polynomials of Class 9 NCERT comprise of very clear cut theory for students to do more practice that needs to solve questions given in the model. Chapter 2 comprises issues dependent on Factorization of Polynomials and Algebraic Identities and polynomials in one variable, Zeros of a Polynomial. Another significant formula you will comprehend is Remainder Theorem.

Importance of NCERT Class 9 Maths Exemplar Book

The NCERT Class 9 Maths Exemplar book comprises the important questions according to the assessment perspective. These NCERT Exemplar questions and answers for CBSE Class 9 Maths Chapter 2 are planned by the experts of maths who target at keeping each idea of the chapter clear for you. The experts at Vedantu have given you stepwise NCERT Exemplar arrangements so you can comprehend the ideas and score more marks in your CBSE tests.

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FAQs on NCERT Exemplar for Class 9 Maths Chapter 2 - Polynomials (Book Solutions)

1. What are the important tips for students to study in NCERT Class 9 Maths Chapter 2? 

Problems in NCERT Exemplar Class 9 Maths Exemplar Chapter 2 will examine various concepts related to polynomials. In the word 'polynomials', 'poly' signifies 'many' and 'nomials' signifies 'term'. As far as mathematics is concerned, a polynomial contains factors that are referred to as coefficients and indeterminates as well. A coefficient includes various activities like addition, subtraction, no negative number type of factors just as multiplication. These are some important tips that Class 9 students should keep in mind while preparing for their Maths board examinations. 

2. What do Polynomial and Constant Polynomials refer to in Class 9 Maths chapter 2 solutions? 

The reference is as follows:

Polynomials : 

Polynomials allude to the expressions that contain at least one term alongside a non-zero coefficient. Polynomials can have more than one term. In a polynomial, every single articulation is referred to as a term.

Constant Polynomials : 

Whereas, At the point when the real numbers are showcased as polynomials like 367 are additionally polynomials that don't contain any factors, these are known as constant polynomials. Apart from this, a constant polynomial which refers to 0  is known as a zero polynomial. 

3. What is Polynomial Function in NCERT Class 9 maths chapter 2?

A polynomial function is a capacity that includes just no negative numbers or just the positive numbers types of any factor in a situation like a quadratic condition, cubic condition, and so on. You can also consider a polynomial function as a polynomial articulation that is characterized by its degree. You can address a polynomial as p{x}. There are many polynomial functions that depend on the level of the polynomial and some of them are: 

Zero polynomial function

Quadratic polynomial function

Linear polynomial function

4. How many questions are there in each exercise of NCERT Exemplar for Class 9 maths chapter 2? 

The first exercise of NCERT Exemplar problems with answers for class 9 maths chapter 2 starts with the questions in which you need to distinguish if the polynomial is in 1 variable or not.

In the second exercise, you need to observe the values of the polynomials in various questions as these questions really look at your insightful abilities.

In activities 2.3.and 2.4, you will be checked assuming you can find, you will be able to distinguish in case a polynomial had the given factor.

Practice 2.5 will make you use the characters in the questions and if you will be able to factorize each question given.

5. Why should we use NCERT Exemplar Solutions Maths Chapter 2 – Polynomials By Vedantu? 

Vedantu f urnishes you with the ideal NCERT Exemplar problems and answers for the class 9 maths chapter to guarantee that you can have the simplest access to the quality study material. We completely consider the significance of a subject and the topic that will be discussed. All the NCERT Exemplar problems with solutions guarantee that you can, without much of a stretch, revise the important points while you have your CBSE tests. Our team is sufficiently proficient to give you the best solutions free of cost!

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CBSE Important Questions for Class 9 Maths are available in Printable format for Free Download.Here you may find NCERT Important Questions and Extra Questions for Class 9 Mathematics chapter wise with answers also. These questions will act as chapter wise test papers for Class 9 Mathematics. These Important Questions for Class 9 Mathematics are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Board Examinations

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Maths Topics to be covered for Class 9

  • Real Numbers
  • Polynomials
  • Coordinate Geometry
  • Linear Equations in Two Variables
  • Introduction to Euclid's Geometry
  • Lines and Angles
  • Heron’s Formula
  • Probability
  • Quadrilaterals
  • Area of Parallelogram and Triangles
  • Constructions
  • Surface Areas and Volumes

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Importance of Question Bank for Exam Preparation?

There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows

  • Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
  • CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.
  • Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.
  • Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.

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Class 9 Polynomials Case Study Questions - Podar International

Test: Polynomials- Case Based Type Questions - Class 9 MCQ

10 questions mcq test - test: polynomials- case based type questions, beti bachao, beti padhao (bbbp) is a personal campaign of the government of india that aims to generate awareness and improve the efficiency of welfare services intended for girls. in a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on beti bachao, beti padhao. q. if in the group, there are 10 teachers and 58 girls, then what is the number of boys.

No. of teachers = x + y = 10

⇒ (x + y) 2 = (10) 2

⇒ x 2 + y 2 + 2xy = 100

[Since (a + b) 2 = a 2 + b 2 + 2ab]

No. of students = (x 2 + y 2 ) = 58

⇒ 58 + 2xy = 100

⇒ 2xy = 100 – 58

⇒ xy = 42/2

Now, since (x + y) 3 = [x 3 + y 3 + 3xy(x + y)]

⇒ (10) 3 = [x 3 + y 3 + 3 × 21(10)]

⇒ 1000 = (x 3 + y 3 + 630)

⇒ 1000 – 630 = (x 3 + y 3 )

⇒ (x 3 + y 3 ) = 370

class 9 maths case study questions with solutions polynomials

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls. In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao. Q. Which is the correct identity?

  • A. (a + b) 2 = a 2 + b 2 – 2ab
  • B. (a + b) 2 = a 2 + b 2 + 2ab
  • C. (a + b) 2 = a 2 – b 2 – 2ab
  • D. All are correct.

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls. In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao. Q. Which mathematical concept is used here?

  • A. Linear equations
  • B. Triangles
  • C. Polynomials

In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls.

class 9 maths case study questions with solutions polynomials

In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao.

Q. (x – y) 3 =

(x 2 – y 2 – 3xy(x – y))

(x 3 – y 3 – 2xy(x – y)

(x 3 – y 3 – 3xy(x – y)

(x 3 – y 3 – 3xyx – y)

We know that (x - y) 3 can be written as

(x - y)(x - y)(x - y)

We know that (x - y)(x - y) can be multiplied and written as

= x 2 - xy - yx + y 2 (x - y)

= x 2 - 2xy + y 2 (x - y)

= x 3 - 2x 2 y + xy 2 - yx 2 + 2xy 2 - y 3

= x 3 – y 3 – 2xy(x – y)

class 9 maths case study questions with solutions polynomials

Q. Using part (D), find (x 2 – y 2 ) if x – y = 23.

Also, x + y = 10

x 2 – y 2 = (x + y)(x – y)

National Association For The Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and productivity.

class 9 maths case study questions with solutions polynomials

Q. Find the amount donated by Ravi.

class 9 maths case study questions with solutions polynomials

Q. (x – y) 3 =

  • A. x 3 – y 3 – 3xy
  • B. x 3 – y 3 – 3xy(x – y)
  • C. x 3 – y 3 – 3xy(x + y)
  • D. x 3 – y 3

x 2 - xy - yx + y 2 (x - y)

= x 3 – y 3 – 3xy(x – y)

class 9 maths case study questions with solutions polynomials

Q. (x + a)(x + b) = x 2 + ................ x + ab

class 9 maths case study questions with solutions polynomials

Q. Which mathematical concept is involved in the above situation?

  • A. Polynomial
  • C. Lines and angles
  • D. Triangle

class 9 maths case study questions with solutions polynomials

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class 9 maths case study questions with solutions polynomials

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CBSE Class 9 Maths Case Study Questions PDF Download

Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.

class 9 maths case study questions with solutions polynomials

Case study questions play a pivotal role in enhancing students’ problem-solving skills. By presenting real-life scenarios, these questions encourage students to think beyond textbook formulas and apply mathematical concepts to practical situations. This approach not only strengthens their understanding of mathematical concepts but also develops their analytical thinking abilities.

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CBSE Class 9th MATHS: Chapterwise Case Study Questions

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Class 9 Maths Case Study Questions, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

Class 9 Maths Case Study Questions

Chapterwise Case Study Questions of Class 9 Maths

  • Case Study Questions for Chapter 1 Number System
  • Case Study Questions for Chapter 2 Polynomials
  • Case Study Questions for Chapter 3 Coordinate Geometry
  • Case Study Questions for Chapter 4 Linear Equations in Two Variables
  • Case Study Questions for Chapter 5 Introduction to Euclid’s Geometry
  • Case Study Questions for Chapter 6 Lines and Angles
  • Case Study Questions for Chapter 7 Triangles
  • Case Study Questions for Chapter 8 Quadrilaterals
  • Case Study Questions for Chapter 9 Areas of Parallelograms and Triangles
  • Case Study Questions for Chapter 10 Circles
  • Case Study Questions for Chapter 11 Constructions
  • Case Study Questions for Chapter 12 Heron’s Formula
  • Case Study Questions for Chapter 13 Surface Area and Volumes
  • Case Study Questions for Chapter 14 Statistics
  • Case Study Questions for Chapter 15 Probability

Checkout: Class 9 Science Case Study Questions

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The above  Class 9 Maths Case Study Question s will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Study Questions have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

Class 9 Maths Syllabus 2023-24

class 9 maths case study questions with solutions polynomials

UNIT I: NUMBER SYSTEMS

1. REAL NUMBERS (18 Periods)

1. Review of representation of natural numbers, integers, and rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.

2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.

3. Definition of nth root of a real number.

4. Rationalization (with precise meaning) of real numbers of the type

jagran josh

(and their combinations) where x and y are natural number and a and b are integers.

5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

UNIT II: ALGEBRA

1. POLYNOMIALS (26 Periods)

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Motivate and State the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Verification of identities:

RELATED STORIES

jagran josh

and their use in factorization of polynomials.

2. LINEAR EQUATIONS IN TWO VARIABLES (16 Periods)

Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c=0.Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.

UNIT III: COORDINATE GEOMETRY COORDINATE GEOMETRY (7 Periods)

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations.

UNIT IV: GEOMETRY

1. INTRODUCTION TO EUCLID’S GEOMETRY (7 Periods)

History – Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example: (Axiom)

1. Given two distinct points, there exists one and only one line through them. (Theorem)

2. (Prove) Two distinct lines cannot have more than one point in common.

2. LINES AND ANGLES (15 Periods)

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180O and the converse.

2. (Prove) If two lines intersect, vertically opposite angles are equal.

3. (Motivate) Lines which are parallel to a given line are parallel.

3. TRIANGLES (22 Periods)

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).

2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).

3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).

4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)

5. (Prove) The angles opposite to equal sides of a triangle are equal.

6. (Motivate) The sides opposite to equal angles of a triangle are equal.

4. QUADRILATERALS (13 Periods)

1. (Prove) The diagonal divides a parallelogram into two congruent triangles.

2. (Motivate) In a parallelogram opposite sides are equal, and conversely.

3. (Motivate) In a parallelogram opposite angles are equal, and conversely.

4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.

5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.

6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

5. CIRCLES (17 Periods)

1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.

2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.

4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

5. (Motivate) Angles in the same segment of a circle are equal.

6. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

UNIT V: MENSURATION 1.

1. AREAS (5 Periods)

Area of a triangle using Heron’s formula (without proof)

2. SURFACE AREAS AND VOLUMES (17 Periods)

Surface areas and volumes of spheres (including hemispheres) and right circular cones.

UNIT VI: STATISTICS & PROBABILITY

STATISTICS (15 Periods)

 Bar graphs, histograms (with varying base lengths), and frequency polygons.

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Benefits of Practicing CBSE Class 9 Maths Case Study Questions

Regular practice of CBSE Class 9 Maths case study questions offers several benefits to students. Some of the key advantages include:

  • Deeper Understanding : Case study questions foster a deeper understanding of mathematical concepts by connecting them to real-world scenarios. This improves retention and comprehension.
  • Practical Application : Students learn to apply mathematical concepts to practical situations, preparing them for real-life problem-solving beyond the classroom.
  • Critical Thinking : Case study questions require students to think critically, analyze data, and devise appropriate solutions. This nurtures their critical thinking abilities, which are valuable in various academic and professional domains.
  • Exam Readiness : By practicing case study questions, students become familiar with the question format and gain confidence in their problem-solving abilities. This enhances their readiness for CBSE Class 9 Maths exams.
  • Holistic Development: Solving case study questions cultivates not only mathematical skills but also essential life skills like analytical thinking, decision-making, and effective communication.

Tips to Solve CBSE Class 9 Maths Case Study Questions Effectively

Solving case study questions can be challenging, but with the right approach, you can excel. Here are some tips to enhance your problem-solving skills:

  • Read the case study thoroughly and understand the problem statement before attempting to solve it.
  • Identify the relevant data and extract the necessary information for your solution.
  • Break down complex problems into smaller, manageable parts to simplify the solution process.
  • Apply the appropriate mathematical concepts and formulas, ensuring a solid understanding of their principles.
  • Clearly communicate your solution approach, including the steps followed, calculations made, and reasoning behind your choices.
  • Practice regularly to familiarize yourself with different types of case study questions and enhance your problem-solving speed.Class 9 Maths Case Study Questions

Remember, solving case study questions is not just about finding the correct answer but also about demonstrating a logical and systematic approach. Now, let’s explore some resources that can aid your preparation for CBSE Class 9 Maths case study questions.

Q1. Are case study questions included in the Class 9 Maths Case Study Questions syllabus?

Yes, case study questions are an integral part of the CBSE Class 9 Maths syllabus. They are designed to enhance problem-solving skills and encourage the application of mathematical concepts to real-life scenarios.

Q2. How can solving case study questions benefit students ?

Solving case study questions enhances students’ problem-solving skills, analytical thinking, and decision-making abilities. It also bridges the gap between theoretical knowledge and practical application, making mathematics more relevant and engaging.

Q3. How do case study questions help in exam preparation?

Case study questions help in exam preparation by familiarizing students with the question format, improving analytical thinking skills, and developing a systematic approach to problem-solving. Regular practice of case study questions enhances exam readiness and boosts confidence in solving such questions.

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Case Study Questions for Class 9 Maths Chapter 12 Herons Formula

Case study questions for class 9 maths chapter 9 areas of parallelograms and triangles, case study questions for class 9 maths chapter 6 lines and angles, case study questions for class 9 maths chapter 7 triangles, case study questions for class 9 maths chapter 5 introduction to euclid’s geometry, case study and passage based questions for class 9 maths chapter 14 statistics, case study questions for class 9 maths chapter 1 real numbers, case study questions for class 9 maths chapter 4 linear equations in two variables, case study questions for class 9 maths chapter 3 coordinate geometry, case study questions for class 9 maths chapter 15 probability, case study questions for class 9 maths chapter 13 surface area and volume, case study questions for class 9 maths chapter 10 circles, case study questions for class 9 maths chapter 9 quadrilaterals, case study questions for class 9 maths chapter 2 polynomials.

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Important Questions Class 9 Maths Chapter 2

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class 9 maths case study questions with solutions polynomials

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Important Questions Class 9 Mathematics Chapter 2- Polynomials.

Mathematics Chapter 2 of Class 9  is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.

Extramarks is one of the leading online learning platforms trusted by lakhs of students and educators across the country.  Students can register on the Extramarks website to access our study materials, including NCERT solutions, CBSE sample papers, revision notes, etc. Our in-house mathematics faculty experts have developed these study materials s by referring to the CBSE curriculum.

Our Mathematics experts believe that students must practice questions regularly to perform  better in exams. For this purpose, they have prepared the Important Questions Class 9 Mathematics Chapter 2 to help students get access to questions from all the topics of the Polynomials. The questions are followed by their step-by-step answers, which will further help students to revise the chapter. The questions are curated from various sources such as the NCERT textbook and exemplar book, CBSE past years’ question papers, and other reference materials. 

Extramarks believes in incorporating joyful learning experiences through its own repository.

To get access to our comprehensive study solutions, students can register on Extramarks platform. Along with the Important Questions Class 9 Mathematics Chapter 2, students will be able to access other study resources such as NCERT solutions, chapter-wise notes, CBSE revision notes, CBSE mock tests, etc.

Important Questions Class 9 Mathematics Chapter 2 – With Solutions

Our in-house Mathematics faculty experts  have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared  illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.

Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:

Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.

Answer 1: Consider the equation 3x + 2y = 12

Now, square both sides:

(3x + 2y)² = 12²

=> 9x² + 12xy + 4y² = 144

=>9x² + 4y² = 144 – 12xy

From the questions, xy = 6

9x² + 4y² = 144 – 72

Thus, the value of 9x² + 4y² = 72

Question 2:Evaluate the following using suitable identity

Answer 2: We can write 102 as 100+2

Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)

(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

Question 3:Without any actual division, prove that the following 2x⁴

 – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

[Hint: Factorise x² – 3x + 2]

Answer 3: x²-3x+2

x(x-2)-1(x-2)

Therefore,(x-2)(x-1)are the factors.

Considering (x-2),

Then, p(x) becomes,

p(x)=2x⁴-5x³+2x²-x+2

p(2)=2(2)⁴-5(2)³+2(2)²-2+2

Therefore, (x-2) is a factor.

Considering (x-1),

p(1)=2(1)⁴-5(1)³+2(1)²-1+2

Therefore, (x-1) is a factor.

Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case

(i) p(x) = 2x³+x²–2x–1, g(x) = x+1

Answer 4 :p(x) = 2x³+x²–2x–1, g(x) = x+1

∴Zero of g(x) is -1.

p(−1) = 2(−1)³+(−1)²–2(−1)–1

∴By the given factor theorem, g(x) is a factor of p(x).

Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.

Answer 5: An example of a monomial having a degree of 82 = x⁸²

An example of a binomial having a required degree of 99 = x⁹⁹ + 7

Question 6: If the two  x – 2 and x – ½ are the given factors of px ²

 + 5 x + r , show that p = r .

Answer 6: Given, f(x) = px²+5x+r and factors are x-2, x – ½

Substituting x = 2 in place of the equation, we get

f(x) = px²+5x+r

f(2) = p(2)²+5(2)+r=0

= 4p + 10 + r = 0 … eq.(i)

Substituting x = ½ in place of the equation, we get,

f( ½ ) = p( ½ )² + 5( ½ ) + r =0

= p/4 + 5/2 + r = 0

= p + 10 + 4r = 0 … eq(ii)

On solving eq(i) and eq(ii),

4p + r = – 10 and p + 4r = – 10

 the RHS of both equations are the same,

4p + r = p + 4r

Hence Proved.

Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.

(i) – 7 + x

(iii) – ? ³

(iv) 1 – y – ? ³

(v) x – ? ³ + ?⁴

(vi) 1 + x + ?²

Answer 7: (i) – 7 + x

The degree of – 7 + x is 1.

Hence, it is a linear polynomial.

The degree of 6y is 1.

Therefore, it is a linear polynomial.

We know that the degree of – ? ³ is 3.

Therefore, it is a cubic polynomial.

We know that the degree of 1 – y – ? ³ is 3.

We know that the degree of x – ? ³ + ?⁴ is 4.

Therefore, it is a quartic polynomial.

We know that the degree of 1 + x + ?² is 2.

Therefore, it is a quadratic polynomial.

We know that the degree of -6?² is 2.

We know that -13 is a constant.

Therefore, it is a constant polynomial.

We know that the degree of –p is 1.

Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.

Answer 8 : Let the polynomial be f(x) = 5x – 4x² + 3

Now, for x = 2,

f(2) = 5(2) – 4(2)² + 3

=> f(2) = 10 – 16 + 3 = –3

Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1)² + 3

=> f(–1) = –5 –4 + 3 = -6

The value of the polynomial 5x – 4x² + 3 at x = -1 is -6.

Question 9:Expanding each of the following, using all the suitable identities:

(i) (x+2y+4z)²

(ii) (2x−y+z)²

(iii) (−2x+3y+2z)²

(iv) (3a –7b–c)²

(v) (–2x+5y–3z)²

Answer 9: (i) (x+2y+4z)²

Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx

Here, x = x

(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)

= x²+4y²+16z²+4xy+16yz+8xz

Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx

Here, x = 2x

(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)

= 4x²+y²+z²–4xy–2yz+4xz

Here, x = −2x

(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)

= 4x²+9y²+4z²–12xy+12yz–8xz

Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx

Here, x = 3a

(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)

= 9a² + 49b² + c²– 42ab+14bc–6ca

Here, x = –2x

(–2x+5y–3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 × 5y× – 3z)+(2×–3z ×–2x)

= 4x²+25y² +9z²– 20xy–30yz+12zx

Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.

Answer 10: Zero of the polynomial,

Hence, zero of g(z) = – 2a

Let p(z) = az³+4z²+3z-4

Now, substituting the given value of z = 3 in p(z), we get,

p(3) = a (3)³ + 4 (3)² + 3 (3) – 4

⇒p(3) = 27a+36+9-4

⇒p(3) = 27a+41

Let h(z) = z³-4z+a

Now, by substituting the value of z = 3 in h(z), we get,

h(3) = (3)³-4(3)+a

⇒h(3) = 27-12+a

⇒h(3) = 15+a

As per the question,

The two polynomials, p(z) and h(z), leave the same remainder when divided by z-3

So, h(3)=p(3)

⇒15+a = 27a+41

⇒15-41 = 27a – a

Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12. 

Answer 11: A rea of rectangle = 25x² – 35x + 12

We know the area of a rectangle = length × breadth

So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.

25x² – 35x + 12 = 25x² – 15x – 20x + 12

=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)

=> 25x² – 35x + 12 = (5x – 3)(5x – 4)

Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).

So, the perimeter = 2(length + breadth)

Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]

                                                                                 = 2(5x – 3 + 5x – 4)

                                                                                = 2(10x – 7) 

                                                                                = 20x – 14

Hence, the perimeter of the rectangle = 20x – 14

Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz

Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx

We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²

2x²+y²+8z²–2√2xy+4√2yz–8xz

= (-√2x)²+(y)²+(2√2z)²+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)

= (−√2x+y+2√2z)²

= (−√2x+y+2√2z)(−√2x+y+2√2z)

Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.

Answer 13: According to the question,

Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a

⟹ x + 2a = 0

Hence, zero of g(x) = – 2a

As per the factor theorem,

If g(x) is a factor of p(x), then p( – 2a) = 0

So, substituting the value of x in p(x), we get,

p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0

⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0

⟹ – 2a = – 3

Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1

Answer 14: Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)

From the question, x² + y² + z²= 83 and x + y + z = 15

152 = 83 + 2(xy + yz + xz)

=> 225 – 83 = 2(xy + yz + xz)

Or, xy + yz + xz = 142/2 = 71

Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),

x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))

x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71

So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)

=> x ³ + y ³ + z ³ – 3xyz = 15 × 12

Or, x ³ + y ³ + z ³ – 3xyz = 180

Question 15:Verify that:

(i) x³+y³ = (x+y)(x²–xy+y²)

(ii) x³–y³ = (x–y)(x²+xy+y²)

Answer 15: (i) x³+y³ = (x+y)(x²–xy+y²)

                     We know that (x+y)³= x³+y³+3xy(x+y)

                      ⇒ x³+y³ = (x+y)³–3xy(x+y)

                      ⇒ x³+y³ = (x+y)[(x+y)²–3xy]

                     Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]

                     ⇒ x³+y³ = (x+y)(x²+y²–xy)

                   (ii) x³–y³ = (x–y)(x²+xy+y²) 

                   We know that (x–y)³ = x³–y³–3xy(x–y)

                    ⇒ x³−y³ = (x–y)³+3xy(x–y)

                    ⇒ x³−y³ = (x–y)[(x–y)²+3xy]

                  Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]

                    ⇒ x³+y³ = (x–y)(x²+y²+xy)

Question 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?

Answer 16: According to the question,

Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2

⟹ x + 2 = 0

Hence, zero of g(x) = – 2

if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.

Thus, substituting the value of x in p(x), we obtain,

p( – 2) = 0

⟹ ( – 2)³ – 2m( – 2)² + 16 = 0

⟹ 0 – 8 – 8m + 16 = 0

Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.

Answer 17: We know that,

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)

Given, a + b + c = 15 and a² + b² + c² = 83

From (ii), we have

152 = 83 + 2(ab + bc + ca)

⇒ 225 – 83 = 2(ab + bc + ca)

⇒ 142/2 = ab + bc + ca

⇒ ab + bc + ca = 71

Now, (i) can be written as

a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]

a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.

Question 18: Factorise: 27x³+y³+z³–9xyz

Answer 18: The expression27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)

27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)

We know that x³+y³+³–3xyz = (x+y+z)(x²+y²+z²–xy –yz–zx)

= (3x+y+z)[(3x)²+y²+z²–3xy–yz–3xz]

= (3x+y+z)(9x²+y²+²–3xy–yz–3xz)

Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).

Answer 19: Given, (x – 1/x) = 4

Squaring both sides, we get,

(x – 1/x)² = 16

⇒ x² – 2.x.1/x + 1/x² = 16

⇒ x² – 2 + 1/x² = 16

⇒ x² + 1/x² = 16 + 2 = 18

∴ (x² + 1/x²) = 18 ….(i)

Again, squaring both sides of (i), we get

(x² + 1/x²)² = 324

⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324

⇒ x⁴ + 2 + 1/x⁴ = 324

⇒ x⁴ + 1/x⁴ = 324 – 2 = 322

∴ (x⁴ + 1/x⁴) = 322.

Question 20: Factorise

Answer 20: The expression 64m³–343n³ can be written as (4m)³–(7n)³

64m³–343n³ =(4m)³–(7n)³

We know that x³–y³ = (x–y)(x²+xy+y²)

64m³–343n³ = (4m)³–(7n)³

= (4m-7n)[(4m)²+(4m)(7n)+(7n)²]

= (4m-7n)(16m²+28mn+49n² )

Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.

Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.

p(2) = 2(2)³ + a(2)² + 2 + b = 0

⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)

p(½) = 2(½)³ + a(½)² + (½) + b = 0

⇒ a + 4b = –3 ….(ii)

On solving (i) and (ii), we get a = 5 and b = –2.

Hence, a = 5 and b = –2.

Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.

Answer 22: According to the question,

Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1

zero of g(p) ⇒ g(p) = 0

Therefore, zero of g(x) = 1

We know that,

According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero

h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0

⟹ g (p) is a factor of h(p).

Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1

Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1

As per the factor theorem, if g (p) is a factor of h(p),

Then h(1) = 0

⟹ (1) ¹¹ – 1 = 0

Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1

Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).

Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.

By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).

Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.

∴ g(x) is not a factor of p(x).

Question 24:Prove that:

x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]

Answer 24: We know that,

x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)

⇒ x³+y³+z³–3xyz = (1/2)(x+y+z)[2(x²+y²+z²–xy–yz–xz)]

= (1/2)(x+y+z)(2×2+2y²+²–2xy–2yz–2xz)

= (1/2)(x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]

= (1/2)(x+y+z)[(x–y)²+(y–z)²+(z–x)²]

Question 25: Find out which of the following polynomials has x – 2 a factor:

(i) 3?² + 6?−24.

(ii) 4?² + ?−2.

Answer 25: (i) According to the question,

Let p(x) =3?² + 6?−24 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

Hence, zero of g(x) = 2

Thus, substituting the value of x in p(x), we get,

p(2) = 3(2)² + 6 (2) – 24

= 12 + 12 – 24

 the remainder = zero,

We can derive that,

g(x) = x – 2 is factor of p(x) = 3?² + 6?−24

(ii) According to the question,

Let p(x) = 4?² + ?−2 and g(x) = x – 2

p(2) = 4(2)² + 2−2

Since the remainder ≠ zero,

We can say that,

g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2

Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.

Answer 26 : x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)

= (x + 1/x)² – 2(x + 1/x)

= (x + 1/x)(x + 1/x – 2).

Question 27: Factorise

8a³+b³+12a²b+6ab²

Answer 27: The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²

8a³+b³+12a²b+6ab² = (2a)³+b³+3(2a)²b+3(2a)(b)²

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.

Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where

(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1

(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3

(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1

(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?

Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1

Here zero of g(x) = – 1

By applying the remainder theorem

P(x) divided by g(x) = p( – 1)

P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0

Therefore, the remainder = 0

(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3

Here zero of g(x) = 3

By applying the remainder theorem p(x) divided by g(x) = p(3)

p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62

Therefore, the remainder = 62

(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1

Here zero of g(x) = ½

By applying the remainder theorem p(x) divided by g(x) = p (½)

P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3

           = 4/8 – 12/4 + 14/2 – 3

           = ½ + 1

           = 3/2

Hence, the remainder = 3/2

so, zero of g(x) = 2/3

By applying the remainder theorem p(x) divided by g(x) = p(2/3)

p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4

Therefore, the remainder = – 136/27

Question 29:Factorise x² – 1 – 2a – a².

Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)

= x² – (1 + a)²

= [x – (1 – a)][x + 1 + a]

= (x – 1 – a)(x + 1 + a)

∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).

Question 30:Evaluate the following using suitable identity

Answer 30: We can write 99 as 1000–2

Using identity,(x–y)³ = x³ –y³ –3xy(x–y)

(998)³ =(1000–2)³ 

=(1000)³ –2³ –(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

Question 31: Find the zeroes of the polynomial:

p(?)= (? –2)² −(? + 2)² 

Answer 31: p(x) = (? –2)² −(? + 2)² 

Zero of the polynomial p(x) = 0

Hence, we get,

⇒ (x–2)² −(x + 2)² = 0

Expanding using the identity, a² – b² = (a – b) (a + b)

⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0

⇒ 2x ( – 4) = 0

Therefore, the zero of the polynomial = 0

Benefits Of Solving Important Questions Class 9 Mathematics Chapter 2

Consistently solving questions is a vital element of mastering   Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter. 

A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are: 

  • Class 9 Mathematics Chapter 2 important questions provide details about the types of questions that may be expected in the exams, which increases their confidence in achieving a high grade. . 
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Q.1 By actual division, find the quotient and the remainder when x 5 + 1 is divided by x 1

Marks: 3 Ans

x 4 + x 3 + x 2 + x + 1 x 1 x 5 + 1 x 5 x 4 + x 4 + 1 x 4 x 3 + x 3 + 1 x 3 x 2 + ¯ x 2 + 1 x 2 x + x + 1 x 1 + 2 ¯ ¯ ¯ ¯ Quotient : x 4 + x 3 + x 2 + x + 1 Remainder : 2

Q.2 Find the value of k if x 5 is a factor of kx 2 + 3x + 7.

Marks: 2 Ans

Zero of x 5 is 5 as x 5 = 0 gives x = 5 . p(x) = kx 2 + 3 x + 7 p ( 5 ) = 0 25 k + 15 + 7 = 0 25 k + 22 = 0 k = 22 25

Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x 2 + y 2 + z 2 .

Given : x + y + z = 6 and xy + yz + zx = 11 Squaring both sides , we get x + y + z 2 = 6 2 x 2 + y 2 + z 2 + 2 xy + 2 yz + 2 zx = 36 x 2 + y 2 + z 2 + 2 xy + yz + zx = 36 x 2 + y 2 + z 2 + 2 11 = 36 Since xy + yz + zx = 11 x 2 + y 2 + z 2 + 22 = 36 x 2 + y 2 + z 2 = 36 22 = 14

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Faqs (frequently asked questions), 1. what are the four types of polynomials.

The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.

2. Where can I get important questions for Class 9 Mathematics Chapter 2 online?

On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.

3. What are the important chapters in Class 9 Mathematics?

The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.

 All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:

  • Chapter – Number Systems
  • Chapter – Polynomials
  • Chapter – Coordinate Geometry
  • Chapter – Linear Equations In Two Variables
  • Chapter – Introduction To Euclid’s Geometry
  • Chapter – Lines And Angles
  • Chapter – Triangles
  • Chapter – Quadrilaterals
  • Chapter – Areas Of Parallelograms And Triangles
  • Chapter – Circles
  • Chapter – Constructions
  • Chapter – Heron’s Formula
  • Chapter – Surface Areas And Volumes
  • Chapter – Statistics
  • Chapter – Probability

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 2 Polynomials NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.1

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1 00001

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2 00001

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3 00001

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4 00001

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5 00001

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials – Topic Discussion

Below we have listed the topics that have been discussed in this chapter. As this is one of the important topics in maths, It comes under the unit – Algebra which has a weightage of 20 marks in class 9 maths board exams.

  • Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
  • Zeroes of a Polynomial
  • Real Numbers and their Decimal Expansions
  • Representing Real Numbers on the Number Line Operations on Real Numbers
  • Laws of Exponents for Real Numbers.

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  4. Polynomials

  5. CLASS 10TH MATH CHAPTER 2 CASE STUDY// POLYNOMIALS CASE STUDY QUESTIONS

  6. Class 9 Maths Polynomials One Shot Revision

COMMENTS

  1. Class 9 Maths Case Study Questions of Chapter 2 Polynomials PDF

    Show Answer Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621 Show Answer The shares of Ankur and Ranjan invested individually are (a) (2x + 1), (2x + 5) (b) (2x + 3), (x + 1) (c) (x + 1), (x + 3) (d) None of these Show Answer Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic

  2. Case Study Questions for Class 9 Maths Chapter 2 Polynomials

    Some of them are given below. Answer them. (i) Which one of the following is not a polynomial? (a) 4x 2 + 2x - 1 (b) y+ (3/y) (c) x 3 - 1 (d) y 2 + 5y + 1 (ii) The polynomial of the type ax2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial

  3. CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with

    CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions. Each question has five sub-questions, each followed by four options and one correct answer.

  4. CBSE Class 9 Mathematics Case Study Questions

    myCBSEguide App Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes. Install Now If you're looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter.

  5. Case Study Questions for Class 9 Maths

    Well, you're in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques.

  6. CBSE Class 9 Maths Polynomials Case Study Questions

    TopperLearning provides a complete collection of case studies for CBSE Class 9 Maths Polynomials chapter. Improve your understanding of biological concepts and develop problem-solving skills with expert advice.

  7. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

    Algebraic Identities Students can refer to the NCERT Solutions for Class 9 while solving exercise problems and preparing for their Class 9 Maths exams. NCERT Class 9 Maths Chapter 2 - Polynomials Summary NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths.

  8. Polynomials Class 9

    Updated for new NCERT - 2023-24 Edition. Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Answers to each and every question is explained in an easy to understand way, with videos of all the questions. In this chapter, we will learn. What is a Polynomial.

  9. Subjective Case Based Questions

    Subjective Case Based Questions -🔥Polynomials Class 9 Maths Chapter 2 | CBSE Board Exam 2022 Preparation | NCERT Solutions | #VedantuClass9and10 Are all wor... CBSE Exam, class 10

  10. CBSE Class 9 Maths Chapter 2

    The important questions of ch 2 Maths class 9 are carefully designed under the CBSE board's rules' strict guidance. To perform well in mathematics, academic success is practice; the students must efficiently practice the polynomials class 9 important questions. To prevent any issues or mistakes in the important questions for class 9 maths ...

  11. Important Questions CBSE Class 9 Maths Chapter 2 Polynomial

    Some important questions from polynomials are given below with solutions. These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently. 1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively. Solution:

  12. CBSE Case Study Questions Class 9 Maths Chapter 2 Polynomials PDF

    Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares. Coefficient of x2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12 Show Answer Total amount invested by both, if x = 1000 is

  13. Case Study Questions Class 9 Chapter 2

    Welcome to our Class 9 Maths exam preparation series for the academic year 2023-24! In this video, we dive deep into Case Study Questions from Chapter 2: Pol...

  14. Polynomial Class 9 Chapter 2 Notes, Examples & Important Questions

    x 2 + x + 1 To know more about polynomials, click here. Term In the polynomial, each expression is called a term . Suppose x 2 + 5x + 2 is polynomial, then the expressions x 2, 5x, and 2 are the terms of the polynomial. Coefficient Each term of the polynomial has a coefficient.

  15. NCERT Exemplar for Class 9 Maths Chapter 2

    Free PDF download of NCERT Exemplar for Class 9 Maths Chapter 2 - Polynomials solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2 - Polynomials exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations. Download free Class 9 Maths to amp up ...

  16. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

    Ex 2.1 Class 9 Maths Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x 2 - 3x + 7 (ii) y 2 + √2 (iii) 3 √t + t√2 (iv) y+ 2 y (v) x 10 + y 3 +t 50 Solution: (i) We have 4x 2 - 3x + 7 = 4x 2 - 3x + 7x 0 It is a polynomial in one variable i.e., x

  17. Class 9 Polynomials Case Study Questions

    7 Views 6 Downloads Class 9 Polynomials Case Study Questions - Podar International School Class 9 Mathematics Questions with answers in PDF format for free Class 9 Maths Question Papers with answers Class 9 Podar International School Mathematics CBSE 2024 Download Paper Related Papers

  18. Test: Polynomials- Case Based Type Questions

    Detailed Solution for Test: Polynomials- Case Based Type Questions - Question 1 No. of teachers = x + y = 10 ⇒ (x + y) 2 = (10) 2 ⇒ x 2 + y 2 + 2xy = 100 [Since (a + b) 2 = a 2 + b 2 + 2ab] No. of students = (x 2 + y 2) = 58 ⇒ 58 + 2xy = 100 ⇒ 2xy = 100 - 58 ⇒ 2xy = 42 ⇒ xy = 42/2 ⇒ xy = 21 Now, since (x + y) 3 = [x 3 + y 3 + 3xy (x + y)]

  19. CBSE Class 9 Maths Case Study Questions PDF Download

    by experts Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.

  20. Category: Case Study Questions for Class 9 Maths

    Case Study Questions for Class 9 Maths Archives - Gurukul of Excellence. Q-104, Sector-38, Gurgaon, Haryana 07065827902 [email protected] M-F: 8 AM- 8 PM; Weekend: 10 AM-2 PM.

  21. Important Questions Class 9 Maths Chapter 2

    Download Important Questions Class 9 Maths Chapter 2 - Polynomials free PDF prepared by subject experts as per the latest NCERT books to score marks in exam ... Important Questions Class 9 Mathematics Chapter 2- Polynomials. Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning "many ...

  22. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

    Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

  23. Polynomials Class 9 Extra Questions Maths Chapter 2 with Solutions Answers

    Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. Extra Questions for Class 9 Maths Polynomials with Answers Solutions. ... CA Foundation Business Economics Study Material - Elasticity of Demand ...