Chapter 2 Polynomial Exercise 2.2 Class 10 Solutions PDF
Neha Tanna21 Nov, 2025
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Chapter 2 Polynomial Exercise 2.2 Class 10 Solutions: NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2 help students understand how the zeros of a polynomial relate to its coefficients. These polynomial exercise 2.2 class 10 solutions provide clear, step-by-step explanations for linear, quadratic, and cubic polynomials.
Created as per the CBSE syllabus, these solutions make learning easier, improve problem-solving skills, and boost exam preparation. Using Chapter 2 polynomials class 10 exercise 2.2 solutions, students can practice efficiently, verify answers, and gain confidence in mastering polynomial concepts.
Class 10 Maths Chapter 2 Ex 2.2 Polynomials Questions and Solutions
Below is the NCERT Solutions for Class 10 Maths Chapter 2 Ex 2.2 Polynomials. These step-by-step solutions for polynomial exercise 2.2 class 10 help students understand zeros, coefficients, and graphical representations clearly for easy practice and exam preparation.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solutions:
(i) x 2 –2x –8
⇒ x 2 – 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x 2 –2x–8 are (4, -2) Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x 2 ) Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x 2 )(ii) 4s 2 –4s+1
⇒4s 2 –2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of the polynomial equation 4s 2 –4s+1 are (1/2, 1/2) Sum of zeroes = (½)+(1/2) = 1 = -(-4)/4 = -(Coefficient of s)/(Coefficient of s 2 ) Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s 2 )(iii) 6x 2 –3–7x
⇒6x 2 –7x–3 = 6x 2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3) Therefore, zeroes of the polynomial equation 6x 2 –3–7x are (-1/3, 3/2) Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2 ) Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x 2 )(iv) 4u 2 +8u
⇒ 4u(u+2) Therefore, zeroes of the polynomial equation 4u 2 + 8u are (0, -2) Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u 2 ) Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u 2 )(v) t 2 –15
⇒ t 2 = 15 or t = ±√15 Therefore, zeroes of the polynomial equation t 2 –15 are (√15, -√15) Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t 2 ) Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t 2 )(vi) 3x 2 –x–4
⇒ 3x 2 –4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1) Therefore, zeroes of the polynomial equation 3x 2 – x – 4 are (4/3, -1) Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x 2 ) Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x 2 )2. Find a quadratic polynomial, each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4 , -1
Solution:
From the formulas of sum and product of zeroes, we know, Sum of zeroes = α+β Product of zeroes = α β Sum of zeroes = α+β = 1/4 Product of zeroes = α β = -1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x +αβ = 0
x 2 –(1/4)x +(-1) = 0
4x 2 –x-4 = 0
Thus, 4x 2 –x–4 is the quadratic polynomial.(ii) √2, 1/3
Solution:
Sum of zeroes = α + β =√2. Product of zeroes = α β = 1/3 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x +αβ = 0
x 2 –( √2 )x + (1/3) = 0
3x 2 -3√2x+1 = 0
Thus, 3x 2 -3√2x+1 is the quadratic polynomial.(iii) 0, √5
Solution:
Given, Sum of zeroes = α+β = 0 Product of zeroes = α β = √5 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x +αβ = 0
x 2 –(0)x + √5 = 0
Thus, x 2 +√5 is the quadratic polynomial.(iv) 1, 1
Solution:
Given, Sum of zeroes = α+β = 1 Product of zeroes = α β = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x +αβ = 0
x 2 –x+1 = 0
Thus, x 2 –x+1 is the quadratic polynomial.(v) -1/4, 1/4
Solution:
Given, Sum of zeroes = α+β = -1/4, Product of zeroes = α β = 1/4 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x +αβ = 0
x 2 –(-1/4)x +(1/4) = 0
4x 2 +x+1 = 0
Thus, 4x 2 +x+1 is the quadratic polynomial.(vi) 4, 1
Solution:
Given, Sum of zeroes = α+β =4, Product of zeroes = αβ = 1 ∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly asx 2 –(α+β)x+αβ = 0
x 2 –4x+1 = 0
Thus, x 2 –4x+1 is the quadratic polynomial.Exercise 2.2 of polynomials of class 10 PDF
These NCERT Solutions for Class 10 Maths Chapter 2 Ex 2.2 explain the relationship between the zeros of polynomials and their coefficients in a simple, step-by-step manner. Designed as per the CBSE syllabus, they help students understand concepts clearly, practice effectively, and prepare confidently for exams. Click below to access the PDF for easy study and revision.
NCERT Solutions for Class 10 Maths PDF
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