NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.2)

Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x²–2x –8
Sol: ⇒ x²– 4x+2x–8
= x(x–4)+2(x–4)
= (x-4)(x+2)
Therefore, zeroes of polynomial equation x²–2x–8 are (4, -2)
Sum of zeroes
= 4–2 = 2
= -(-2)/1 = -(Coefficient of x)/(Coefficient of x²)
Product of zeroes
= 4×(-2)
= -8
=-(8)/1 = (Constant term)/(Coefficient of x²)

(ii) 4s²–4s+1
Sol: ⇒ 4s²–2s–2s+1
= 2s(2s–1)–1(2s-1)
= (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s²–4s+1 are (1/2, 1/2)
Sum of zeroes
= (½)+(1/2)
= 1
= -4/4 = -(Coefficient of s)/(Coefficient of s²)
Product of zeros
= (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s² )

(iii) 6x²–3–7x
Sol: ⇒ 6x²–7x–3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) +1(2x – 3)
= (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x²–3–7x are (-1/3, 3/2)
Sum of zeroes
= -(1/3)+(3/2)
= (7/6) = -(Coefficient of x)/(Coefficient of x²)
Product of zeroes
= -(1/3)×(3/2)
= -(3/6) = (Constant term) /(Coefficient of x² )

(iv) 4u²+8u
Sol: ⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u² + 8u are (0, -2).
Sum of zeroes
= 0+(-2)
= -2
= -(8/4) = -(Coefficient of u)/(Coefficient of u²)
Product of zeroes
= 0×-2
= 0
= 0/4 = (Constant term)/(Coefficient of u² )

(v) t²–15
Sol: ⇒ t² = 15 or t = ±√15
Therefore, zeroes of polynomial equation t² –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t²)
Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t² )

(vi) 3x²–x–4
Sol: ⇒ 3x²–4x+3x–4
= x(3x-4)+1(3x-4)
= (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation3x² – x – 4 are (4/3, -1)
Sum of zeroes
= (4/3)+(-1)
= (1/3)
= -(-1/3) = -(Coefficient of x) / (Coefficient of x²)
Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x² )

Q2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
Sol:
From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α+β
Product of zeroes = αβ
Sum of zeroes = α+β = 1/4
Product of zeroes = αβ = -1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:
x²–(α+β)x +αβ = 0
x²–(1/4)x +(-1) = 0
4x²–x-4 = 0
Thus,4x²–x–4 is the quadratic polynomial.

(ii)√2, 1/3
Sol:
Sum of zeroes = α + β =√2
Product of zeroes = αβ = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:
x²–(α+β)x +αβ = 0
x² –(√2)x + (1/3) = 0
3x²-3√2x+1 = 0
Thus, 3x²-3√2x+1 is the quadratic polynomial.

(iii) 0, √5
Sol:
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 as:
x²–(α+β)x +αβ = 0
x²–(0)x +√5= 0
Thus, x²+√5 is the quadratic polynomial.

(iv) 1, 1
Sol:
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 as:
x²–(α+β)x +αβ = 0
x²–x+1 = 0
Thus, x²–x+1 is the quadratic polynomial.

(v) -1/4, 1/4
Sol:
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 as:
x²–(α+β)x +αβ = 0
x²–(-1/4)x +(1/4) = 0
4x²+x+1 = 0
Thus,4x²+x+1 is the quadratic polynomial.

(vi) 4, 1
Sol:
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 as:
x²–(α+β)x+αβ = 0
x²–4x+1 = 0
Thus, x²–4x+1 is the quadratic polynomial.

Check out the NCERT Solutions of all the exercises of Polynomials: 

Exercise 2.1. NCERT Solutions: Polynomials

Exercise 2.3 NCERT Solutions: Polynomials

Exercise 2.4 NCERT Solutions: Polynomials