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NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.3

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NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 - Free PDF Download

The NCERT Solutions for Class 9 Maths Exercise 2.3 Polynomials provides complete solutions to the problems in the Exercise. These NCERT Solutions are intended to assist students with the CBSE Class 9 examination. Students should thoroughly study this NCERT solution in order to solve all types of questions based on arithmetic progression. By completing these practice questions with the NCERT Maths Solutions Chapter 2 Exercise 2.3 Class 9, you will be better prepared to understand all of the different types of questions that may be asked in the Class 9 exams.

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Table of Content
1. NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 - Free PDF Download
2. Glance on NCERT Solutions Maths Chapter 2 Polynomials Exercise 2.3 Class 9 | Vedantu
3. Access NCERT Solution for Class 9 Maths Chapter 2 Polynomials Exercise 2.3
4. Conclusion
5. NCERT Solutions for Class 9 Maths Chapter 2 Exercises
6. CBSE Class 9 Maths Chapter 2 Polynomials Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 9 Maths
8. Important Study Materials for CBSE Class 9 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 2 Polynomials Exercise 2.3 Class 9 | Vedantu

  • Chapter 2 of the Class 9 NCERT Maths book focuses on Polynomials, and Exercise 2.3 specifically deals with the concept of zeros of a polynomial. 

  • Understanding Zeros of a Polynomial: A zero of a polynomial p(x) is a value ‘a’ such that p(a)=0.

  • This exercise involves finding the zeros of various polynomials and verifying the results by substituting back into the polynomial equation.

  • The exercise often involves factorizing the polynomial to find its zeros.

  • For quadratic polynomials, this typically involves expressing the polynomial in the form (x−a)(x−b) and solving for x.

  • After determining the potential zeros, verify the solutions by substituting them back into the original polynomial equation to check if it equals zero.

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NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.3
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Access NCERT Solution for Class 9 Maths Chapter 2 Polynomials Exercise 2.3

1. Determine which of the following polynomials has\[\left( {x + 1} \right)\] a factor:

i. \[{x^3} + {x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^3} + {x^2} + x + 1\]

Now, for \[x =  - 1\]

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

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

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

Therefore, by the Factor Theorem, \[x + 1\] is a factor of \[{x^3} + {x^2} + x + 1\].

ii. \[{x^4} + {x^3} + {x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^4} + {x^3} + {x^2} + x + 1\]

Now, for \[x =  - 1\]

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

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

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

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of \[{x^4} + {x^3} + {x^2} + x + 1\].

iii. \[{x^4} + 3{x^3} + 3{x^2} + x + 1\]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^4} + 3{x^3} + 3{x^2} + x + 1\]

Now, for \[x =  - 1\]

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

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

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

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of \[{x^4} + 3{x^3} + 3{x^2} + x + 1\].

iv. \[{x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \]

Ans: We know that, 

Zero of \[x + 1\] is \[ - 1\]

Given that, 

\[p\left( x \right) = {x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \]

Now, for \[x =  - 1\]

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

\[p\left( { - 1} \right) =  - 1 + 1 + 2 - \sqrt 2  + \sqrt 2 \]

\[p\left( { - 1} \right) = 2 + 2\sqrt 2 \]

Therefore, by the Factor Theorem, \[x + 1\] is not a factor of\[{x^3} + {x^2} - \left( {2 + \sqrt 2 } \right)x + \sqrt 2 \].

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

i. \[p\left( x \right) = 2{x^3} + {x^2} - 2x - 1\], \[g\left( x \right) = x + 1\]

Ans: Given that, 

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

\[g\left( x \right) = x + 1\]

We know that, 

Zero of \[g\left( x \right)\] is \[ - 1\]

Now,

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

\[p\left( { - 1} \right) =  - 2 + 1 + 2 - 1\]

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

Therefore, \[g\left( x \right) = x + 1\] is a factor of\[p\left( x \right) = 2{x^3} + {x^2} - 2x - 1\].


ii. \[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\], \[g\left( x \right) = x + 2\]

Ans: Given that, 

\[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\]

\[g\left( x \right) = x + 2\]

We know that, 

Zero of \[g\left( x \right)\] is \[ - 2\]

Now,

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

\[p\left( { - 2} \right) =  - 8 + 12 - 6 + 1\]

\[p\left( { - 2} \right) =  - 1\]

Therefore, \[g\left( x \right) = x + 2\] is not a factor of \[p\left( x \right) = {x^3} + 3{x^2} + 3x + 1\].

iii. \[p\left( x \right) = {x^3} - 4{x^2} + x + 6\], \[g\left( x \right) = x - 3\]

Ans: Given that, 

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

\[g\left( x \right) = x - 3\]

We know that, 

Zero of \[g\left( x \right)\] is \[3\]

Now,

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

\[p\left( 3 \right) = 27 - 36 + 3 + 6\]

\[p\left( 3 \right) = 0\]

Therefore, \[g\left( x \right) = x - 3\] is a factor of \[p\left( x \right) = {x^3} - 4{x^2} + x + 6\].


3. Find the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right)\] in each of the following cases:

i. \[p\left( x \right) = {x^2} + x + k\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = {x^2} + x + k\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

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

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

\[ \Rightarrow 1 + 1 + k = 0\]

\[ \Rightarrow k =  - 2\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = {x^2} + x + k\] is \[ - 2\].


ii. \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

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

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

\[ \Rightarrow 2 + k + \sqrt 2  = 0\]

\[ \Rightarrow k =  - \left( {2 + \sqrt 2 } \right)\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = 2{x^2} + kx + \sqrt 2 \] is \[ - \left( {2 + \sqrt 2 } \right)\].


iii. \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

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

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

\[ \Rightarrow k - \sqrt 2  + 1 = 0\]

\[ \Rightarrow k = \sqrt 2  - 1\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} - \sqrt 2 x + 1\] is $({\sqrt 2}-1)$.


iv. \[p\left( x \right) = k{x^2} + 3x + k\]

Ans: Given that \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} + 3x + k\]

Thus, \[1\] is the zero of the given \[p\left( x \right)\] 

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

\[ \Rightarrow p\left( 1 \right) = k{\left( 1 \right)^2} + 3\left( 1 \right) + k = 0\]

\[ \Rightarrow k - 3 + k = 0\]

\[ \Rightarrow k = \frac{3}{2}\]

Therefore, the value of \[k\], if \[x - 1\] is a factor of \[p\left( x \right) = k{x^2} + 3x + k\] is \[\frac{3}{2}\].

4. Factorise:

i. \[12{x^2} - 7x + 1\]

Ans: Given that, 

\[p\left( x \right) = 12{x^2} - 7x + 1\]

Splitting the middle term

\[ \Rightarrow 12{x^2} - 4x + 3x + 1\]

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

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

Therefore,  \[12{x^2} - 4x + 3x + 1 = \left( {3x - 1} \right)\left( {4x - 1} \right)\].

ii. \[2{x^2} + 7x + 3\]

Ans: Given that, 

\[p\left( x \right) = 2{x^2} + 7x + 3\]

Splitting the middle term

\[ \Rightarrow 2{x^2} + x + 6x + 3\]

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

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

Therefore, \[2{x^2} + x + 6x + 3 = \left( {2x + 1} \right)\left( {x + 3} \right)\].

iii. \[6{x^2} + 5x - 6\]

Ans: Given that, 

\[p\left( x \right) = 6{x^2} + 5x - 6\]

Splitting the middle term

\[ \Rightarrow 6{x^2} + 9x - 4x - 6\]

\[ \Rightarrow 3x\left( {2x + 3} \right) - 2\left( {2x + 3} \right)\]

\[ \Rightarrow \left( {2x + 3} \right)\left( {3x - 2} \right)\]

Therefore,  \[6{x^2} + 9x - 4x - 6 = \left( {2x + 3} \right)\left( {3x - 2} \right)\].

iv. \[3{x^2} - x - 4\]

Ans: Given that, 

\[p\left( x \right) = 3{x^2} - x - 4\]

Splitting the middle term

\[ \Rightarrow 3{x^2} - 4x + 3x - 4\]

\[ \Rightarrow x\left( {3x - 4} \right) + 1\left( {3x - 4} \right)\]

\[ \Rightarrow \left( {3x - 4} \right)\left( {x + 1} \right)\]

Therefore,  \[3{x^2} - 4x + 3x - 4 = \left( {3x - 4} \right)\left( {x + 1} \right)\].

5. Factorise:

i. \[{x^3} - 2{x^2} - x + 2\]

Ans: Given that, 

\[p\left( x \right) = {x^3} - 2{x^2} - x + 2\]

Rearranging the above,

\[ \Rightarrow {x^3} - x - 2{x^2} + 2\]

\[ \Rightarrow x\left( {{x^2} - 1} \right) - 2\left( {{x^2} - 1} \right)\]

\[ \Rightarrow \left( {{x^2} - 1} \right)\left( {x - 2} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 2} \right)\]

Therefore, \[{x^3} - 2{x^2} - x + 2 = \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 2} \right)\].

ii. \[{x^3} - 3{x^2} - 9x - 5\]

Ans: Given that, 

\[p\left( x \right) = {x^3} - 3{x^2} - 9x - 5\]

\[ \Rightarrow {x^3} + {x^2} - 4{x^2} - 4x - 5x - 5\]

\[ \Rightarrow {x^2}\left( {x + 1} \right) - 4x\left( {x + 1} \right) - 5\left( {x + 1} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} - 4x - 5} \right)\]

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

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

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

Therefore, \[{x^3} - 3{x^2} - 9x - 5 = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x - 5} \right)\].

iii. \[{x^3} + 13{x^2} + 32x + 20\]

Ans: Given that, 

\[p\left( x \right) = {x^3} + 13{x^2} + 32x + 20\]

\[ \Rightarrow {x^3} + {x^2} + 12{x^2} + 12x + 20x + 20\]

\[ \Rightarrow {x^2}\left( {x + 1} \right) + 12x\left( {x + 1} \right) + 20\left( {x + 1} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 12x + 20} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left( {{x^2} + 2x + 10x + 20} \right)\]

\[ \Rightarrow \left( {x + 1} \right)\left[ {x\left( {x + 2} \right) + 10\left( {x + 2} \right)} \right]\]

\[ \Rightarrow \left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right)\] 

Therefore, \[{x^3} + 13{x^2} + 32x + 20 = \left( {x + 1} \right)\left( {x + 10} \right)\left( {x + 2} \right)\].

iv. \[2{y^3} + {y^2} - 2y - 1\]

Ans: Given that, 

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

\[ \Rightarrow 2{y^3} - 2{y^2} + 3{y^2} - 3y + y - 1\]

\[ \Rightarrow 2{y^2}\left( {y - 1} \right) + 3y\left( {y - 1} \right) + 1\left( {y - 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2{y^2} + 3y + 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2{y^2} + 2y + y + 1} \right)\]

\[ \Rightarrow \left( {y - 1} \right)\left[ {2y\left( {y + 1} \right) + 1\left( {y + 1} \right)} \right]\]

\[ \Rightarrow \left( {y - 1} \right)\left( {2y + 1} \right)\left( {y + 1} \right)\]

Therefore, \[2{y^3} + {y^2} - 2y - 1 = \left( {y - 1} \right)\left( {2y + 1} \right)\left( {y + 1} \right)\].


Conclusion

Class 9 Maths Ex 2.3 of Chapter 2 - Polynomials, is crucial for a solid foundation in math. Understanding the concept of factoring quadratic expressions is a key takeaway. Vedantu's NCERT solutions can guide you in conquering quadratic equations using factorization. Regular practice with NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in AP-related problems


NCERT Solutions for Class 9 Maths Chapter 2 Exercises

Exercise

Number of Questions

Exercise 2.1

5 Questions & Solutions

Exercise 2.2

4 Questions & Solutions

Exercise 2.4

5 Questions & Solutions


CBSE Class 9 Maths Chapter 2 Polynomials Other Study Materials



Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Materials for CBSE Class 9 Maths

FAQs on NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.3

1. Are the answers provided by Vedantu, sufficient to attain accurate marks?

Our answers are made to the point and they are drafted to aid you from the exam portion of the view. Answers to the exercising questions are certainly given with examples and they are 100% curate. Our answers will make your learning easy for the exam as they are fitted to be compatible with the tips given with the assistance of using CBSE maths Syllabus and NCERT Book.


Our answers will assist you in advancing a conceptual basis with all the principal ideas in a comprehensible language. The exercising covers all of the vital topics and subtopics of the chapter which might occur for your Class 9 maths exams. You also can clear all of your doubts from here and in this manner, you can definitely perform well in your Class 9 maths exam.

2. Give a brief description of the topic

Polynomial is acquired from the word “poly” which means “many” and the word “nominal” refers to “term”. In Math subject, a polynomial expression consists of variables which are also known as coefficients and indeterminates. The coefficients require the operations of addition, subtraction, non-negative integer exponents of variables and multiplication. 


Polynomials are utterances with one or more phrases with a non-zero coefficient. A polynomial can also have one or more than one number of terms. In the form of a polynomial, each utterance in it is known as a term.  Suppose x2 + 4x+ 2 is polynomial, then the expressions x2, 4x, and 2 are the terms of the polynomial.  Each term of the polynomial has a coefficient. For example, if 4x+1 is the polynomial, then the coefficient of x is 4.

3. What are the topics that are covered in this Chapter?

Polynomial is an algebraic expression which includes constants, variables and exponents. It is the expression where the variables have just certain elemental powers. The topics that are covered in this chapter are:

  • Introduction

  • Polynomials in One Variable

  • Zeros of Polynomials

  • Remainder Theorem

  • Factorisation of Polynomials

  • Algebraic Identities

Polynomial expressions are algebraic equations which will have two or more terms with the equal variables of different exponents. This is one of the notable topics for class 9 Mathematics, which students require to learn in order to gain profound knowledge about complex algebraic expressions. Mathematics is one of those subjects that serve as a foundation of numbers, analysis, figures and logic.

4. What are the important terms of a polynomial?

  • A term might either be a variable or a single digit or it also could be a sequence of the variable with digits.

  • The degree of the polynomial is the most distinguished power of the variable in a polynomial.

  • A polynomial of degree 1 will be termed as a linear polynomial.

  • A polynomial of degree 2 will be described as a quadratic polynomial.

  • A polynomial of degree 3 is termed as a cubic polynomial.

  • A polynomial of 1 term is termed as a monomial.

  • A polynomial of 2 terms is termed as binomial.

  • A polynomial of 3 terms is termed as a trinomial.

  • A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0, where a is also termed as root of the equation p(x) = 0.

  • A linear polynomial in one variable holds an individual zero, a polynomial of a non-zero constant has no zero, and every real character is a zero of the zero polynomial.

5. How many questions are there in NCERT Solutions in Chapter 2 Class Maths 9 Exercise 2.4?

There are five questions in NCERT Solutions of Class 9 Maths Chapter 2 Exercise 2.4. To practice all these questions, you can download the NCERT solutions prepared by professionals at Vedantu. These solutions help students understand all the concepts and revise before the exams. Click on NCERT Solutions for Class 9 Maths Chapter 2 to download the PDF of NCERT Solutions Class 8 Maths Chapter 2 to study the solutions offline. These solutions are available free of cost on Vedantu (vedantu.com). You can download it using the Vedantu app as well.

6. Do NCERT Solutions prepared by Vedantu also provide important questions for Class 9 Maths Chapter 2?

Yes. The NCERT solutions provided by Vedantu also contain important questions. You can download them and use them to prepare for your exams. As they follow the CBSE syllabus and are prepared by a panel of experts, these solutions help students achieve their perfect scores. Click on Important Questions for CBSE Class 9 Maths Chapter 2  to look at the important questions and answers for Chapter 2 of Class 9 Maths. You can also register for online classes on Vedantu (vedantu.com) to make studying more fun and interesting.

7. Do I need to practice all the questions in Exercise 2.4 of Chapter 2 Class 9 Maths?

Yes, you need to practice all the questions and solutions provided in the NCERT solutions. Maths can be difficult to handle but once you start practising and understanding the concepts, it will be the easiest subject of all. Click on NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.4) to download the PDF of NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 provided by Vedantu.

8.  What is the best solution book for Chapter 2 Maths Class 9?

The best solution book one can opt for CBSE Maths is NCERT Solutions. Vedantu provides the best study material for Class 9 Maths Chapter 2, created by experts. They provide a series of problems and solutions to help students clear their doubts and prepare for the exams. Click on NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.4) to download the NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 to ace your exams.

9. Is Chapter 2 Maths Class 9 difficult?

No. Once you understand concepts and clear your doubts, you will easily solve all the problems in exams. To get more guidance on Chapter 2 Class 9 Maths, click on NCERT Solutions for Class 9 Maths Chapter 2 Polynomials to view the PDF of NCERT Solutions prepared by experts at Vedantu. These guides will help you comprehend the concepts easily and can be used as reference material to revise before the exam.