# Ncert maths solutions for class 10 chapter 2 Polynomials

# 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) x ^{2}–2x –8**

**⇒ **x^{2}– 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 x^{2})

= -(-2)/1 = 2

Product of zeroes = 4×(-2) = -8

Again product of zeroes using formula = (Constant term)/(Coefficient of x^{2})

=-(8)/1 = -8

**verify**

**(ii) 4s ^{2}–4s+1**

⇒4s^{2}–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 s^{2})

= -(-4)/4 = 1

Product of zeros = (1/2)×(1/2) = 1/4

Again product of the zeroes using formula = (Constant term)/(Coefficient of s^{2 })

= 1/4

**verify**

**(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 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 x^{2})

= -(-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^{2 })

= -3/6

**(iv) 4u ^{2}+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 u^{2})

= -(8/4) = -2

Product of zeroes = 0×-2 = 0

Again product of the zeroes using formula = (Constant term)/(Coefficient of u^{2 })

= 0/4

**verify**

**(v) t ^{2}–15**

⇒ t^{2} = 15 or t = ±√15

Therefore, zeroes of 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 })

**verify**

**(vi) 3x ^{2}–x–4**

⇒ 3x^{2}–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 x^{2})

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

Product of zeroes=(4/3)×(-1) = (-4/3)

Again product of the zeroes using formula = (Constant term) /(Coefficient of x^{2 })

= (-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:-

x^{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

The quadratic polynomial equation can be written as:-

x^{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

The quadratic polynomial equation can be written as:-

x^{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

The quadratic polynomial equation can be written as:-

x^{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

Then the quadratic polynomial equation can be written directly as:-

x^{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

Then the quadratic polynomial equation can be written directly as:-

x^{2}–(α+β)x+αβ = 0

⇒ x^{2}–4x+1 = 0

Thus, x^{2}–4x+1 is the quadratic polynomial.

## Ncert solution Class 10

### Chapter 1: Real Number

#### Ncert maths solution class 10 chapter 1.1

#### Ncert maths solution class 10 chapter 1.2

#### Chapter 4:Quadratic Equation

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