Ncert solutions for class 10 maths (chapter 2, exercise 2.3)

12/26/21, 10:05 PM NCERT Solutions for Class 10 Maths (Chapter 2, Exercise 2.3)
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NCERT Solutions for Class 10 Maths (Chapter 2,
Exercise 2.3)
December 25, 2021 by Abdur Rohman
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)
Find Formula

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 by the polynomial
and find the quotient and remainder in each of the
following.
i. ,
Solution:
Dividing the polynomial by ,
p(x)
g(x)
p(x) = x −
3
3x +
2
5x − 3 g(x) = x −
2
2
p(x) g(x)

Here, the Quotient is and the Remainder is .
ii. ,
Solution:
Here, is in standard form but is not in the standard form.
The standard form of would be, .
Now, dividing the polynomial by ,
x − 3 7x − 9
p(x) = x −
4
3x +
2
4x + 5 g(x) = x +
2
1 − x
p(x) g(x)
g(x) g(x) = x −
2
x + 1
p(x) g(x)
x +
2
x − 3 8

iii. ,
Solution:
Here, is in standard form but is not in the standard form.
The standard form of would be, .
Now, dividing the polynomial by ,
Question 2: Check whether the first polynomial is a factor of
the second polynomial by dividing the second polynomial by
the first polynomial.
i. ,
Solution:
Dividing the second polynomial by the first polynomial,
p(x) = x −
4
5x + 6 g(x) = 2 − x2
p(x) g(x)
g(x) g(x) = −x +
2 2
p(x) g(x)
−x −
2
2 −5x + 10
t −
2
3 2t +
4
3t −
3
2t −
2
9t − 12

Here, the remainder is . Therefore, the first polynomial is a factor of the second
polynomial.
ii. ,
Solution:
Here, the remainder is . Therefore, the first polynomial is a factor of the second
polynomial.
iii. ,
Solution:
0
x +
2
3x + 1 3x +
4
5x −
3
7x +
2
2x + 2
0
x −
3
3x + 1 x −
5
4x +
3
x +
2
3x + 1

Here, the remainder is . Therefore, the first polynomial is not a factor of the second
polynomial.
Question 3: Obtain all the zeros of
, if two of its zeros are and .
Solution:
Given, two zeros of the polynomial are, and .
the two factors of the polynomial will be,
and
Multiplying the two factors we get,
Now, is a factor of the polynomial.
is also a factor of the given polynomial.
Dividing by we get
2(
=
 0)
3x +
4
6x −
3
2x −
2
10x − 5

3
5
−

3
5

3
5
−

3
5
∴
x −
(
3
5
) x +
(
3
5
)
x −
x + =
(
3
5
) (
3
5
) x −
2
=
3
5

3
3x −5
2

3
3x −5
2
∴ 3x −
2
5
3x +
4
6x −
3
2x −
2
10x − 5 3x −
2
5

Here, the quotient is and the remainder is .
The quotient can also be written as,
The zeros of the quotient will be, and .
Therefore, all the zeros of the given polynomial are, , , and .
Question 4: On dividing by a polynomial
, the quotient and the remainder were and
, respectively. Find .
Solution:
Let us consider the polynomial as , the quotient as
and the remainder as .
According to the division algorithm for polynomials,
x +
2 2x + 1 0
x +
2 2x + 1 = (x + 1)2
−1 −1

3
5
−

3
5
−1 −1
x −
3 3x +
2 x + 2
g(x) x − 2 −2x +
4 g(x)
p(x) = x −
3
3x +
2
x + 2 q(x) =
x − 2 r(x) = −2x + 4
Dividend = (Divisor × Quotient) + Remainder
⇒ p(x) = g(x) × q(x) + r(x) … … (i)

Putting the values of and in we get,
Now,
Therefore, .
Question 5: Give examples of polynomial
and , which satisfy the division algorithm and
i. deg deg
Solution:
Let,
p(x), q(x) r(x) equation(i)
x −
3 3x +
2 x + 2 = g(x) × (x − 2) + (−2x + 4)
⇒ x −
3 3x +
2 x + 2 = g(x) × (x − 2) − 2x + 4
⇒ x −
3
3x +
2
x + 2 + 2x − 4 = g(x) × (x − 2)
⇒ x −
3
3x +
2
3x − 2 = g(x) × (x − 2)
⇒ g(x) =
x−2
x −3x +3x−2
3 2
g(x) = x −
2 x + 1
p(x), g(x), q(x)
r(x)
p(x) = q(x)

and
Now,
Therefore, the above polynomials satisfy the division algorithm for polynomials and also
deg deg .
ii. deg deg
Solution:
Let,
and
Now,
p(x) = 10x +
2 15x + 20
g(x) = 5
q(x) = 2x +
2 3x + 4
r(x) = 0
p(x) = g(x) × q(x) + r(x)
g(x) × q(x) + r(x) = 5 × (2x +
2
3x + 4) + 0
⇒ g(x) × q(x) + r(x) = 10x +
2 15x + 20 = p(x)
p(x) = q(x) = 2
q(x) = r(x)
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
r(x) = 2x −
2
2x + 1
p(x) = g(x) × q(x) + r(x)

deg deg .
iii. deg
Solution:
Let,
and
Now,
deg .
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.
g(x) × q(x) + r(x) = (x +
3 x +
2 x + 1)(x +
2 x + 1) + 2x −
2 2x + 1
⇒ g(x) × q(x) + r(x) = x +
5 2x +
4 +3x +
3 5x +
2 2 = p(x)
q(x) = r(x) = 2
r(x) = 0
p(x) = 2x +
4 8x +
3 6x +
2 4x + 12
g(x) = x +
4
4x +
3
3x +
2
2x + 1
q(x) = 2
r(x) = 10
p(x) = g(x) × q(x) + r(x)
g(x) × q(x) + r(x) = (x +
4 4x +
3 3x +
2 2x + 1 × 2 + 10
⇒ g(x) × q(x) + r(x) = 2x +
4 8x +
3 6x +
2 4x + 12 = p(x)
r(x) = 0

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List of Exercises in Polynomials, Chapter 2, Class 10 Maths
Exercise 2.1 (NCERT Solutions for Ex. 2.1 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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