NCERT Solutions for Class 10 Maths (Chapter 2, Exercise 2.3)

NCERT Solutions for Class 10 Maths Chapter 2 is provided in this section. The Chapter 2 of Class 10 Mathematics under the NCERT Syllabus is about the Polynomials. As per the Textbook published by NCERT, the chapter discusses the concepts of the Geometrical Meaning of the Zeros of a Polynomial, the Relationship between Zeros and Coefficients of a Polynomial, and the Division Algorithm for Polynomials.

This post contains the step-by-step solutions of Exercise 2.3, Chapter 2 of Class 10 Mathematics designed in the easiest possible way. There is a total of 5 questions in the Ex. 2.3 and all the questions are covered in this section.

Topics Covered in Chapter 2: Polynomials

Serial No.	Section	Topic
1	2.1	Introduction
2	2.2	Geometrical Meaning of the Zeros of a Polynomial
3	2.3	Relationship between Zeros and Coefficients of a Polynomial
4	2.4	Division Algorithm for Polynomials
5	2.5	Summary

NCERT Solutions for Class 10 Maths, Chapter 2 – Polynomials (Ex. 2.3)

NCERT Solutions for Class 10 Maths, Chapter 2 (Polynomials), Exercise 2.3 are described below. The Exercise 2.3 is based on Division Algorithm for Polynomials and each and every question of Ex. 2.3 are solved (step-by-step) below.

Board	CBSE
Textbook (Council)	NCERT (National Council of Educational Research and Training)
Class	Class 10 or Class X
Subject	Mathematics
Chapter Number	Chapter 2
Chapter Name	Polynomials
Exercise Number	Exercise 2.3
Number of Questions	5 Questions

Exercise 2.3

Question 1: Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.

i. p(x)=x^3-3x^2+5x-3, g(x)=x^2-2

Solution:

Dividing the polynomial p(x) by g(x),

Here, the Quotient is x-3 and the Remainder is 7x-9.

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ii. p(x)=x^4-3x^2+4x+5, g(x)=x^2+1-x

Solution:

Here, p(x) is in standard form but g(x) is not in the standard form.

The standard form of g(x) would be, g(x)=x^2-x+1.

Now, dividing the polynomial p(x) by g(x),

Here, the Quotient is x^2+x-3 and the Remainder is 8.

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iii. p(x)=x^4-5x+6, g(x)=2-x^2

Solution:

Here, p(x) is in standard form but g(x) is not in the standard form.

The standard form of g(x) would be, g(x)=-x^2+2.

Now, dividing the polynomial p(x) by g(x),

Here, the Quotient is -x^2-2 and the Remainder is -5x+10.

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Question 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

i. t^2-3, 2t^4+3t^3-2t^2-9t-12

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is 0. Therefore, the first polynomial is a factor of the second polynomial.

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ii. x^2+3x+1, 3x^4+5x^3-7x^2+2x+2

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is 0. Therefore, the first polynomial is a factor of the second polynomial.

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iii. x^3-3x+1, x^5-4x^3+x^2+3x+1

Solution:

Dividing the second polynomial by the first polynomial,

Here, the remainder is 2 (\ne 0). Therefore, the first polynomial is not a factor of the second polynomial.

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Question 3: Obtain all the zeros of 3x^4+6x^3-2x^2-10x-5, if two of its zeros are \sqrt{\frac{5}{3}} and -\sqrt{\frac{5}{3}}.

Solution:

Given, two zeros of the polynomial are, \sqrt{\frac{5}{3}} and -\sqrt{\frac{5}{3}}.

\therefore the two factors of the polynomial will be,

\left(x-\sqrt{\frac{5}{3}}\right) and \left(x+\sqrt{\frac{5}{3}}\right)

Multiplying the two factors we get,

\left(x-\sqrt{\frac{5}{3}}\right) \left(x+\sqrt{\frac{5}{3}}\right)=x^2-\frac{5}{3}=\frac{3x^2-5}{3}

Now, \frac{3x^2-5}{3} is a factor of the polynomial.

\therefore\,3x^2-5 is also a factor of the given polynomial.

Dividing 3x^4+6x^3-2x^2-10x-5 by 3x^2-5 we get

Here, the quotient is x^2+2x+1 and the remainder is 0.

The quotient can also be written as,

x^2+2x+1=(x+1)^2

The zeros of the quotient will be, -1 and -1.

Therefore, all the zeros of the given polynomial are, \sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}, -1 and -1.

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Question 4: On dividing x^3-3x^2+x+2 by a polynomial g(x), the quotient and the remainder were x-2 and -2x+4, respectively. Find g(x).

Solution:

Let us consider the polynomial as p(x)=x^3-3x^2+x+2, the quotient as q(x)=x-2 and the remainder as r(x)=-2x+4.

According to the division algorithm for polynomials,

Dividend=(Divisor\times Quotient)+Remainder

\Rightarrow p(x)=g(x)\times q(x)+r(x)\,\,\,\,\,\,\,\,\,\,……(i)

Putting the values of p(x),\,q(x) and r(x) in equation(i) we get,

x^3-3x^2+x+2=g(x)\times (x-2)+(-2x+4)

\Rightarrow x^3-3x^2+x+2=g(x)\times (x-2)-2x+4

\Rightarrow x^3-3x^2+x+2+2x-4=g(x)\times (x-2)

\Rightarrow x^3-3x^2+3x-2=g(x)\times (x-2)

\Rightarrow g(x)=\frac{x^3-3x^2+3x-2}{x-2}

Now,

Therefore, g(x)=x^2-x+1.

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Question 5: Give examples of polynomial p(x),\,g(x),\,q(x) and r(x), which satisfy the division algorithm and

i. deg p(x)=deg q(x)

Solution:

Let,

p(x)=10x^2+15x+20

g(x)=5

q(x)=2x^2+3x+4 and

r(x)=0

According to the division algorithm for polynomials,

p(x)=g(x)\times q(x)+r(x)

Now,

g(x)\times q(x)+r(x)=5\times (2x^2+3x+4)+0

\Rightarrow g(x)\times q(x)+r(x)=10x^2+15x+20=p(x)

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg p(x)=deg q(x)=2.

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ii. deg q(x)=deg r(x)

Solution:

Let,

p(x)=x^5+2x^4++3x^3+5x^2+2

g(x)=x^3+x^2+x+1

q(x)=x^2+x+1 and

r(x)=2x^2-2x+1

According to the division algorithm for polynomials,

p(x)=g(x)\times q(x)+r(x)

Now,

g(x)\times q(x)+r(x)=(x^3+x^2+x+1)(x^2+x+1)+2x^2-2x+1

\Rightarrow g(x)\times q(x)+r(x)= x^5+2x^4++3x^3+5x^2+2 =p(x)

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg q(x)=deg r(x)=2.

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iii. deg r(x)=0

Solution:

Let,

p(x)=2x^4+8x^3+6x^2+4x+12

g(x)=x^4+4x^3+3x^2+2x+1

q(x)=2 and

r(x)=10

According to the division algorithm for polynomials,

p(x)=g(x)\times q(x)+r(x)

Now,

g(x)\times q(x)+r(x)=(x^4+4x^3+3x^2+2x+1\times 2+10

\Rightarrow g(x)\times q(x)+r(x)= 2x^4+8x^3+6x^2+4x+12=p(x)

Therefore, the above polynomials satisfy the division algorithm for polynomials and also deg r(x)=0.

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NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 PDF Download

The PDF of NCERT solutions for Class 10 Maths Chapter 2, Exercise 2.3 is provided below.

List of Exercises in Polynomials, Chapter 2, Class 10 Maths

• Exercise 2.1 (NCERT Solutions for Ex. 2.1 Class 10 Maths, Chapter 2)
• Exercise 2.2 (NCERT Solutions for Ex. 2.2 Class 10 Maths, Chapter 2)
• Exercise 2.3 (NCERT Solutions for Ex. 2.3 Class 10 Maths, Chapter 2)
• Exercise 2.4 (NCERT Solutions for Ex. 2.4 Class 10 Maths, Chapter 2)

I hope the NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.3 have helped you. The solutions are designed keeping in mind that everyone can understand very easily if previous class concepts are clear.

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