# NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2

## Introduction:

In this exercise/article we will learn about Polynomials. Zeroes of the polynomial. A zero of a polynomial need to not be 0. 0 may be a zero of a polynomial. Every linear polynomial has one and only one zero. A polynomial can have more than one zero.

NCERT Class 9 Maths Chapter 2 Polynomials :

- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.1
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.4
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.5
- Extra Questions Class 9 Maths Chapter 2 Polynomials

**Class 9 Maths Exercise 2.2 (Page-34)**

**Q1. **Find the value of the polynomial 5x - 4x^{2} + 3 at

**(i) **x = 0

**Solution :**

Given 5x - 4x^{2} + 3

Now, 5 × 0 - 4 ( 0 )^{2} + 3 [ put x = 0 ]

= 0 - 4 × 0 + 3

= 0 - 0 + 3 [ because 0 - 0 = 0 ]

= 3

So, answer is 3

**(ii) **x = -1

**Solution :**

Given 5x - 4x^{2} + 3

Now, 5 ( -1 ) - 4 (-1 )^{2} + 3 [ put x = -1 ]

= - 5 - 4 × 1 + 3

= - 5 - 4 + 3

= - 9 + 3

= - 6

So, answer is - 6

**(iii) **x = 2

**Solution :**

Given 5x - 4x^{2} + 3

Now, 5 × 2 - 4 ( 2 )^{2} + 3 [ put x = 2 ]

= 10 - 4 × 4 + 3

= 10 - 16 + 3

= 13 - 16

= - 3

So, answer is - 3

**Q2. **Find p(0), p(1) and p(2) for each of the following polynomials :

**(i) **p(y) = y^{2} - y + 1

**Solution :**

Given y^{2} - y + 1

[ After put y = 0 ]

( 0 )^{2} - 0 + 1

= 0 - 0 + 1 [ because 0 - 0 = 0 ]

= 1

[ After put y = 1 ]

( 1 )^{2} - 1 + 1

= 1 - + 1

= 1

[ After put y = 2 ]

( 2 )^{2} - 2 + 1

= 4 - 2 + 1

= 5 - 2

= 3

After put p(0), p(1) and p(2) we get 1, 1 and 3.

**(ii) **p(t) = 2 + t + 2t^{2} - t^{3}

**Solution :**

Given 2 + t + 2t^{2} - t^{3}

[ After put t = 0 ]

2 + 0 + 2 ( 0 )^{2} - ( 0 )^{3}

= 2 + 0 + 2 × 0 - 0

= 2 + 0 + 0 - 0 [ because 0 - o = o ]

= 2

[ After put t = 1 ]

2 + 1 + 2 ( 1 )^{2} - ( 1 )^{3}

2 + 1 + 2 × 1 - 1

= 2 + 1 + 2 - 1 [ because 1 - 1 = 0 ]

= 4

[ After put t = 2 ]

2 + 2 + 2 ( 2 )^{2} - ( 2 )^{3}

2 + 2 + 2 × 4 - 8

= 2 + 2 + 8 - 8 [ because 8 - 8 = 0 ]

= 4

After put p(0), p(1) and p(2) we get 2, 4 and 4.

**(iii) **p( x ) = x^{3}

**Solution :**

Given x^{3}

[ After put x = 0 ]

( 0 )^{3}

= 0

[ After put x = 1 ]

( 1 )^{3}

= 1

[ After put x = 2 ]

( 2 )^{3}

= 8

After put p(0), p(1) and p(2) we get 0, 1 and 8.

**(iv) **p(x) = ( x - 1 ) ( x + 1 )

**Solution :**

Given ( x - 1 ) ( x + 1 ) [ using identity = ( x - y ) ( x + y ) = x^{2} - y^{2} ]

[ After put x = 0 ]

= ( 0 )^{2} - ( 1 )^{2}

= 0 - 1

= -1

[ After put x = 1 ] [ using identity = ( x - y ) ( x + y ) = x^{2} - y^{2} ]

= ( 1 )^{2} - ( 1 )^{2}

= 1 - 1

= 0

[ After put x = 2 ] [ using identity = ( x - y ) ( x + y ) = x^{2} - y^{2} ]

= ( 2 )^{2} - ( 1 )^{2}

= 4 - 1

= 3

After put p(0), p(1) and p(2) we get -1, 0 and 3.

# NCERT Solutions Class 9 Maths

**Q3. **Verify whether the following are zeroes of the polynomial, indicated against them.

**(i) **p(x) = 3x + 1 , x = \(\displaystyle -\,\frac{1}{3}\)

**Solution :**

Given p(x) = 3x + 1 , x = \(\displaystyle -\,\frac{1}{3}\)

[ After put x = \(\displaystyle -\,\frac{1}{3}\) ]

3 \(\displaystyle \left( {-\,\frac{1}{3}} \right)\) + 1

=

= - 1 + 1

= 0

So, \(\displaystyle -\,\frac{1}{3}\) is a zero of the polynomial 3x + 1 .

**(ii) **p(x) = 5x - \(\displaystyle \pi \) , x = \(\displaystyle \frac{4}{5}\)

**Solution :**

Given p(x) = 5x - \(\displaystyle \pi \) , x = \(\displaystyle \frac{4}{5}\)

[ After put x = \(\displaystyle \frac{4}{5}\) ]

5 × \(\displaystyle \frac{4}{5}\) + \(\displaystyle \pi \)

=

= 4 + \(\displaystyle \frac{4}{5}\)

So, \(\displaystyle \frac{4}{5}\) is a not zero of the polynomial 5x + 1 .

**(iii) **p(x) = x^{2} - 1 , x = 1 , -1

**Solution :**

Given p(x) = x^{2} - 1 , x = 1 , -1

[ After put x = 1 ]

( 1 )^{2} - 1

= 1 - 1

= 0

[ After put x = -1 ]

( - 1 )^{2} - 1

= 1 - 1

= 0

So, 1 and -1 is a zero of the polynomial x^{2} - 1 .

**(iv) **p(x) = ( x + 1 ) ( x - 2 ) , x = -1 , 2

**Solution :**

Given p(x) = ( x + 1 ) ( x - 2 ) , x = -1 , 2

[ After put x = -1 ]

( - 1 + 1 ) ( - 1 - 2 )

= ( 0 ) ( - 3 )

= 0 × - 3

= 0

[ After put x = 2 ]

( 2 + 1 ) ( 2 - 2 )

= ( 3 ) ( 0 )

= 3 × 0

= 0

So, -1 and 2 is a zero of the polynomial ( x + 1 ) ( x - 2 ) .

**(v) **p(x) = x^{2} , x = 0

**Solution :**

Given p(x) = x^{2} , x = 0

[ After put x = 0 ]

= ( 0 )^{2}

= 0

So, 0 is a zero of the polynomial x^{2} .

**(vi) **p(x) = lx + m , x = \(\displaystyle -\,\frac{m}{l}\)

**Solution :**

Given p(x) = lx + m , x = \(\displaystyle -\,\frac{m}{l}\)

[ After put x = \(\displaystyle -\,\frac{m}{l}\) ]

l \(\displaystyle \left( {-\,\frac{m}{l}} \right)\) + m

=

= - m + m

= o

So, \(\displaystyle -\,\frac{m}{l}\) is a zero of the polynomial lx + m .

**(vii) **p(x) = 3x^{2} - 1 , x = \(\displaystyle -\,\frac{1}{{\sqrt{3}}}\,,\,\frac{2}{{\sqrt{3}}}\)

**Solution :**

Given p(x) = 3x^{2} - 1 , x = \(\displaystyle -\,\frac{1}{{\sqrt{3}}}\,,\,\frac{2}{{\sqrt{3}}}\)

[ After put x = \(\displaystyle {-\,\frac{1}{{\sqrt{3}}}}\) ]

3 \(\displaystyle {{\left( {-\,\frac{1}{{\sqrt{3}}}} \right)}^{2}}\) - 1

= 3 × \(\displaystyle \frac{1}{3}\) - 1

=

= 1 - 1

= 0

[ After put x = \(\displaystyle \frac{2}{{\sqrt{3}}}\) ]

3 \(\displaystyle {{\left( {\frac{2}{{\sqrt{3}}}} \right)}^{2}}\) - 1

= 3 × \(\displaystyle \left( {\frac{4}{3}} \right)\) - 1

=

= 4 - 1

= 3

So, \(\displaystyle -\frac{1}{{\sqrt{3}}}\) is a zero but \(\displaystyle \frac{2}{{\sqrt{3}}}\) is not a zero of the polynomial 3x^{2} - 1 .

**(viii) **p(x) = 2x + 1 , x = \(\displaystyle \frac{1}{2}\)

**Solution :**

Given p(x) = 2x + 1 , x = \(\displaystyle \frac{1}{2}\)

[ After put x = \(\displaystyle \frac{1}{2}\) ]

2 \(\displaystyle \left( {\frac{1}{2}} \right)\) + 1

= \(\displaystyle 2\,\times \,\frac{1}{2}\,+\,1\)

=

= 1 + 1

= 2

So, \(\displaystyle \frac{1}{2}\,\) is not a zero of the polynomial 2x + 1 .

**Q4. **Find the zero of the polynomial in each of the following cases :

**(i) **p(x) = x + 5

**Solution :**

Given p(x) = x + 5

[ After put x = 0 ]

x + 5 = 0

x = - 5

So, - 5 is zero of the polynomial x .

**(ii) **p(x) = x - 5

**Solution :**

Given p(x) = x - 5

[ After put x = 0 ]

x - 5 = 0

x = 5

So, 5 is zero of the polynomial x .

**(iii) **p(x) = 2x + 5

**Solution :**

Given p(x) = 2x + 5

2x + 5 = 0

2x = - 5

x = \(\displaystyle -\,\frac{5}{2}\)

So, \(\displaystyle -\,\frac{5}{2}\) is zero of the polynomial x .

**(iv) **p(x) = 3x - 2

**Solution :**

Given p(x) = 3x - 2

3x - 2 = 0

3x = 2

x = \(\displaystyle \frac{2}{3}\)

So, \(\displaystyle \frac{2}{3}\) is zero of the polynomial x .

**(v) **p(x) = 3x

**Solution :**

Given p(x) = 3x

3x = 0

x = - 3

So, - 3 is zero of the polynomial x .

**(vi) **p(x) = ax, a \(\displaystyle \ne \) 0

**Solution :**

Given p(x) = ax, a \(\displaystyle \ne \) 0

ax = 0

x = \(\displaystyle \frac{0}{a}\)

x = 0

So, 0 is zero of the polynomial x .

(vii) p(x) = cx + d, c \(\displaystyle \ne \) 0 , c , d are real numbers.

**Solution :**

Given p(x) = cx + d, c \(\displaystyle \ne \) 0 , c , d are real numbers

cx + d = 0

cx = - d

x = \(\displaystyle -\,\frac{d}{c}\)

So, \(\displaystyle -\,\frac{d}{c}\) is zero of the polynomial x .