NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2

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 :

Class 9 Maths Exercise 2.2 (Page-34)

Q1. Find the value of the polynomial 5x - 4x2 + 3 at

(i) x = 0

Solution :

Given 5x - 4x2 + 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 - 4x2 + 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 - 4x2 + 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) = y2 - y + 1

Solution :

Given y2 - 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 + 2t2 - t3

Solution :

Given 2 + t + 2t2 - t3

[ 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 ) = x3

Solution :

Given x3

[ 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 ) = x2 - y2 ]

[ After put x = 0 ]

= ( 0 )2 - ( 1 )2

= 0 - 1

= -1

[ After put x = 1 ]      [ using identity = ( x - y ) ( x + y ) = x2 - y2 ]

= ( 1 )2 - ( 1 )2

= 1 - 1

= 0

[ After put x = 2 ]            [ using identity = ( x - y ) ( x + y ) = x2 - y2 ]

= ( 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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (i)

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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (ii)

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

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

(iii) p(x) = x2 - 1 ,    x = 1 , -1

Solution :

Given p(x) = x2 - 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 x2 - 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) = x2 ,   x = 0

Solution :

Given p(x) = x2 ,   x = 0

[ After put x = 0 ]

= ( 0 )2

= 0

So, 0 is a zero of the polynomial x2 .

(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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (vi)

= - m + m

= o

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

(vii) p(x) = 3x2 - 1 ,    x = \(\displaystyle -\,\frac{1}{{\sqrt{3}}}\,,\,\frac{2}{{\sqrt{3}}}\)

Solution :

Given p(x) = 3x2 - 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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (vii)

= 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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (vii)

= 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 3x2 - 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

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

= NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2 Q.3, (viii)

= 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 .

 

NCERT Class 9 Maths Chapter 2 Polynomials :

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