Ncert solutions for class 10 maths (chapter 2, exercise 2.4)

12/26/21, 10:06 PM NCERT Solutions for Class 10 Maths (Chapter 2, Exercise 2.4) - Find Formula
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NCERT Solutions for Class 10 Maths (Chapter 2,
Exercise 2.4)
December 26, 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 2.4, Chapter 2 of Class 10
Mathematics designed in the easiest possible way. There is a total of 5 questions in the Ex.
2.4 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. 2.4)
Find Formula

NCERT Solutions for Class 10 Maths, Chapter 2 (Polynomials), Exercise 2.4 are described
below. The Exercise 2.4 is based on Division Algorithm for Polynomials and each and
every question of Ex. 2.4 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 2.4
Number of Questions 5 Questions
Exercise 2.4
Question 1: Verify that the numbers given alongside of the
cubic polynomials below are their zeroes. Also, verify the
relationship between the zeroes and coefficients in each case.
i. ;
Solution:
Let,
Now,
Similarly,
2x +
3
x −
2
5x + 2 , 1, −2
2
1
p(x) = 2x +
3 x −
2 5x + 2
p =
(2
1
) 2 +
(2
1
)
3
−
(2
1
)
2
5 +
(2
1
) 2
p =
(2
1
) +
4
1
−
4
1
+
2
5
2 = 0

Again,
Therefore, and are the zeroes of the given polynomial.
Verifying:
Let,
and
Now,
And
Again
Therefore, the relationship between the zeroes and the coefficients of the polynomial is
verified.
p(1) = 2(1) +
3 (1) −
2 5(1) + 2
p(1) = 2 + 1 − 5 + 2 = 0
p(−2) = 2(−2) +
3
(−2) −
2
5(−2) + 2
p(−2) = −16 + 4 + 10 + 2 = 0
, 1
2
1
−2
α =
2
1
β = 1
γ = −2
α + β + γ = +
2
1
1 − 2
⇒ α + β + γ =
=
2
1+2−4
−
=
2
1
−
Coefficient of x3
Coefficient of x2
αβ + βγ + γα =
(1) +
(2
1
) (1)(−2) + (−2)
(2
1
)
⇒ αβ + βγ + γα = −
2
1
2 − 1 = =
2
1−4−2
=
2
−5

Coefficient of x3
Coefficient of x
αβγ = ×
2
1
1 × (−2)
αβγ = − =
2
2
−
Coefficient of x3
Constant term

ii. ;
Solution:
Let,
Now,
Similarly,
Therefore, and are the zeroes of the given polynomial.
Verifying:
Let,
and
Now,
And,
Again,
x −
3
4x +
2
5x − 2 2, 1, 1
p(x) = x −
3 4x +
2 5x − 2
p(2) = (2) −
3
4(2) +
2
5(2) − 2
p(2) = 8 − 16 + 10 − 2 = 0
p(1) = (1) −
3 4(1) +
2 5(1) − 2
p(1) = 1 − 4 + 5 − 2 = 0
2, 1 1
α = 2
β = 1
γ = 1
α + β + γ = 2 + 1 + 1 = 4 = − =
1
−4
−
Coefficient of x3
Coefficient of x2
αβ + βγ + γα = 2 × 1 + 1 × 1 + 1 × 2 = 2 + 1 + 2 = 5 = =
1
5

Coefficient of x3
Coefficient of x

Therefore, the relationship between the zeroes and the coefficients of the polynomial is
verified.
Question 2: Find a cubic polynomial with the sum, sum of the
product of its zeroes taken two at a time, and the product of its
zeroes as respectively.
Solution:
Let us consider the zeroes of the cubic polynomial be and .
Then, according to the question,
and
Therefore, the required polynomial will be,
Question 3: If the zeroes of the polynomial
are and , then find and .
Solution:
Given, and are the zeros of the polynomial .
Comparing the given polynomial with , we get
αβγ = 2 × 1 × 1 = 2 = − =
1
−2
−
Coefficient of x3
Constant term
2, −7, −14
α, β γ
α + β + γ = 2
αβ + βγ + γα = −7
αβγ = −14
x −
3
(α + β + γ)x +
2
(αβ + βγ + γα)x − (αβγ) = x −
3
2x −
2
7x + 14
x −
3
3x +
2
x + 1
a − b, a a + b a b
(a − b), a (a + b) x −
3 3x +
2 x + 1
Ax +
3
Bx +
2
Cx + D

and
Now, sum of zeros
Putting the values of and ,
Again, the sum of the product of zeroes taken two at a time,
Putting the values of and we get
Therefore, and
NCERT Solutions for Class 10 Maths Chapter 2, Exercise 2.4 PDF
Download
The PDF of NCERT solutions for Class 10 Maths Chapter 2, Exercise 2.4 is provided below.
(Embed)
A = 1, B = −3, C = 1 D = 1
= (a − b) + a + (a + b) = −
A
B
B A
⇒ 3a =
1
−(−3)
⇒ 3a = 3
⇒ a = 1
⇒ a(a − b) + a(a + b) + (a + b)(a − b) =
A
C
C A
⇒ a −
2
ab + a +
2
ab + a −
2
b =
2
1
1
⇒ 3a −
2
b =
2
1
⇒ 3(1) −
2
b =
2
1
⇒ −b =
2 1 − 3
⇒ b =
2 2
⇒ b = ±
2
a = 1 b = ±
2





(Download Link)
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 2.4 have helped you.
The solutions are designed keeping in mind that everyone can understand very easily if
previous class concepts are clear.
If you have any issues/queries related to NCERT solutions for Class 10 Maths, Chapter 2, Ex
2.4, feel free to contact me at contact@findformula.co.in or fill the form here.
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