# Chapter 2 : Polynomials

Ncert maths solutions for class 10 chapter 2 Polynomials , serve as an invaluable companion for students embarking on their mathematical journey. This chapter introduces students to the world of polynomials. NCERT solutions for Chapter 2 are designed to simplify the learning process and foster a genuine appreciation for the elegance of polynomials in mathematics.

## Exercise 2.1 Polynomials

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. ### Solution:

(i) The graph is parallel to x-axis and not cut it at any point hence the number of zeroes of p(x) is 0 .

(ii) The graph intersects the x-axis at only one point hence the number of zeroes of p(x) is 1 .

(iii) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 .

(iv)  The graph intersects the x-axis at two points hence the number of zeroes of p(x) is 2 .

(v) The graph intersects the x-axis at four points hence the number of zeroes of p(x) is 4 .

(vi) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 .

## Exercise 2.2 Polynomials

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solutions:

(i) x2–2x –8

⇒  x2– 4x+2x–8

=   x(x–4)+2(x–4)

=   (x-4)(x+2)

x – 4 = 0 or  x + 2 = 0

⇒  x = 4  or x = -2

Therefore, zeroes of polynomial  = 4, -2

Sum of zeroes = 4–2 = 2 =

Again sum of the zeroes using formula =  -(Coefficient of x)/(Coefficient of x2)

= -(-2)/1 = 2

Product of zeroes = 4×(-2) = -8

Again product of zeroes using formula  = (Constant term)/(Coefficient of x2)

=-(8)/1 = -8

verify

(ii) 4s2–4s+1

⇒4s2–2s–2s+1

= 2s(2s–1)–1(2s-1)

= (2s–1)(2s–1)

Therefore, zeroes of polynomial   =  1/2, 1/2

Sum of zeroes = (½)+(1/2) = 1

Again sum of the zeroes using formula  = -(Coefficient of s)/(Coefficient of s2)

= -(-4)/4 = 1

Product of zeros = (1/2)×(1/2) = 1/4

Again product of the zeroes using formula = (Constant term)/(Coefficient of s)

= 1/4

verify

(iii) 6x2–3–7x

⇒ 6x2–7x–3

= 6x– 9x + 2x – 3

= 3x(2x – 3) +1(2x – 3)

= (3x+1)(2x-3)

Therefore, zeroes of polynomial  =  -1/3,  3/2

Sum of zeroes = -(1/3)+(3/2) = (7/6)

Again sum of the zeroes using formula = -(Coefficient of x)/(Coefficient of x2)

= -(-7)/6 = 7/6

Product of zeroes = -(1/3)×(3/2) = -3/6

Again product of the zeroes using formula = (Constant term) /(Coefficient of x)

=  -3/6

(iv) 4u2+8u

⇒ 4u(u+2)

Therefore, zeroes of polynomial =  0, -2

Sum of zeroes = 0+(-2) = -2

Again sum of the zeroes using formula  = -(Coefficient of u)/(Coefficient of u2)

= -(8/4) = -2

Product of zeroes = 0×-2 = 0

Again product of the zeroes using formula  = (Constant term)/(Coefficient of u)

= 0/4

verify

(v) t2–15

⇒ 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 t)

verify

(vi) 3x2–x–4

⇒ 3x2–4x+3x–4

= x(3x-4)+1(3x-4)

= (3x – 4)(x + 1)

Therefore, zeroes of polynomial  =  (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)

Again sum of the zeroes using formula  = -(Coefficient of x) / (Coefficient of x2)

= -(-1/3) = 1/3

Product of zeroes=(4/3)×(-1) = (-4/3)

Again product of the zeroes using formula = (Constant term) /(Coefficient of x)

=  (-4/3)

verify

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.

(i) 1/4 , -1

Solution:

Let zeroes of the quadratic polynomial

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

The quadratic polynomial equation can be written  as:-

x2–(α+β)x +αβ = 0

⇒ x2–(1/4)x +(-1) = 0

⇒ 4x2–x-4 = 0

(ii)√2, 1/3

Solution:

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

The quadratic polynomial equation can be written  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

Solution:

Given,

Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

The quadratic polynomial equation can be written as:-

x2–(α+β)x +αβ = 0

⇒ x2–(0)x +√5= 0

Thus, x2+√5 is the quadratic polynomial.

(iv) 1, 1

Solution:

Given,

Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

The quadratic polynomial equation can be written as:-

x2–(α+β)x +αβ = 0

⇒ x2–x+1 = 0

Thus, x2–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

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

Solution:

Given,

Sum of zeroes = α+β =4

Product of zeroes = αβ = 1

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.

## Ncert solution Class 10

### Chapter 1: Real Number

#### NCERT Solutions Maths for Class 10 Chapter 4 