NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4

 

Introduction :

In this exercise we will learn about Polynomials. Factor theorem , if p(x) is a polynomial of degree n ≥ 1 and a is any real number then

(i) x - a is a factor of p(x), if p(a) = 0, and

(ii) p(a) = 0, if x - a is a factor of p(x).

This actually follows immediately from the Remainder Theorem, but we shall not prove it here. Using by splitting the middle term.

NCERT Class 9 Maths Chapter 2 Polynomials :

Class 9 Maths Exercise 2.4 (Page-43)

Q1. Determine which of the following polynomials has ( x + 1 ) a Factor :

(i) x3 + x2 + x + 1

Solution :

Let p(x) = x3 + x2 + x + 1

( x + 1 ) is a Factor of p ( x )

So, x + 1 = 0

x = - 1

Using the factor theorem,

When p(x) = x3 + x2 + x + 1

than p( - 1 ) = ( - 1 )3 + ( - 1 )2 + ( - 1 ) + 1

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

= 0

∴ ( x + 1 ) is a factor of x3 + x2 + x + 1 .

(ii) x4 + x3 + x2 + x + 1

Solution :

Let p(x) = x4 + x3 + x2 + x + 1

( x + 1 ) is a Factor of p ( x )

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = x4 + x3 + x2 + x + 1

than p( - 1 ) = ( - 1 )4 + ( - 1 )3 + ( - 1 )2 + ( - 1 ) + 1

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

= 1

∴ ( x + 1 ) is a factor of x4 + x3 + x2 + x + 1 .

(iii) x4 + 3x3 + 3x2 + x + 1

Solution :

Let p(x) = x4 + 3x3 + 3x2 + x + 1

( x + 1 ) is a Factor of p ( x )

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = x4 + 3x3 + 3x2 + x + 1

than p( - 1 ) = ( - 1 )4 + 3 ( - 1 )3 + 3 ( - 1 )2 + ( - 1 ) + 1

= 1 + 3 × ( - 1 ) + 3 × ( 1 ) - 1 + 1

= 1 - 3 + 3 - 1 + 1

= 1

∴ ( x + 1 ) is a factor of x4 + 3x3 + 3x2 + x + 1 .

(iv) x3 - x2 - ( 2 + √2 ) x + √2

Solution :

Let p(x) = x3 - x2 - ( 2 + √2 ) x + √2

( x + 1 ) is a Factor of p ( x )

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = x3 - x2 - ( 2 + √2 ) x + √2

than p ( - 1 ) = ( - 1 )3 - ( - 1 )2 - ( 2 + √2 ) × ( - 1 ) + √2

= - 1 - 1 - ( - 2 - √2 ) + √2

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

= 2√2

∴ ( x + 1 ) is a factor of x3 - x2 - ( 2 + √2 ) x + √2 .

Q2. Use the factor theorem to determine whether g( x ) is a factor of p( x ) in each of the following cases :

(i) p( x ) = 2x3 + x2 - 2x - 1    ,  g( x ) = x + 1

Solution :

Given p(x) = 2x3 + x2 - 2x - 1

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = 2x + x - 2x - 1

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

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

= 0

∴ Yes, g( x ) is a factor of p( x ) .

(ii) p( x ) = x3 + 3x2 + 3x + 1    ,  g( x ) = x + 2

Solution :

Given p(x) = x3 + 3x2 + 3x + 1

So, x + 2 = 0

x = - 2

Using the factor theorem,

when p(x) = x3 + 3x2 + 3x + 1

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

= - 8 + 3 × 4 - 6 + 1

= - 8 + 12 - 6 + 1

= - 14 + 13

= - 1

∴ No, g( x ) is not a factor of p( x ) .

(iii) p( x ) = x3 - 4x2 + x + 6     ,  g( x ) = x - 3

Solution :

Given p(x) = x3 - 4x2 + x + 6

So, x - 3 = 0

x = 3

Using the factor theorem,

when p(x) = x3 - 4x2 + x + 6

p( 3 ) = ( 3 )3 - 4 ( 3 )2 + 3 + 6

= 27 + 4 × 9 + 3 + 6

= 27 - 36 + 3 + 6

= 36 - 36

= 0

∴ Yes, g( x ) is a factor of p( x ) .

 

NCERT Solutions Class 9 Maths Chapter 2

 

Q3. Find the value of k, if x - 1 is a factor of p( x ) in each of the following cases :

(i) p( x ) = x2 + x + k

Solution :

We have p(x) = x2 + x + k

Given x - 1 is factor of p(x)

So,  x - 1 = 0

x = 1

Using the factor theorem,

when p( x ) = x2 + x + k

than p(1) = ( 1 )2 + 1 + k =0

⇒ 1 + 1 + k = 0

⇒ 2 + k = 0

⇒ k = - 2

The value of k = - 2 .

(ii) p( x ) = 2x2 + kx + √2

Solution :

We have p(x) = 2x2 + kx + √2

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = 2x2 + kx + √2

than p( 1 ) = 2 ( 1 )2 + k ( 1 ) + √2 = 0

= 2 × 1 + k + √2 = 0

= 2 + k + √2 = 0

= √2 + k = - 2

= k = - 2 - √2

= k = - ( 2 + √2 )

The value of k = - ( 2 + √2 ) .

(iii) p( x ) = kx2 - √2x + 1

Solution :

We have p(x) = kx2 - √2x + 1

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = kx2 - √2x + 1

than p( 1 ) = k ( 1 )2 - √2 ( 1 ) + 1 = 0

= k - √2 + 1 = 0

= k + 1 = √2

= k = √2 - 1

The value of k = √2 - 1 .

(iv) p( x ) = kx2 - 3x + k

Solution :

We have p(x) = kx2 - 3x + k

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = kx2 - 3x + k

than p( 1 ) = k ( 1 ) - 3 × 1 + k = 0

= k - 3 + k = 0

= 2k - 3 = 0

= 2k = 3

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

The value of k = \(\displaystyle \frac{3}{2}\) .

Q4. Factorise :

(i) 12x2 - 7x + 1

Solution :

We have 12x2 - 7x + 1

[ Using the splitting the middle term method ]

So, 12x2 - 7x + 1

= 12x2 - 4x + 3x + 1

= 3 × 4 × x2 - 4x - 3x + 1   [ find the common ]

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

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

(ii) 2x2 + 7x + 3

Solution :

We have 2x2 + 7x + 3

[ Using the splitting the middle term method ]

So, 2x2 + 7x + 3

= 2x2 + 6x + 1x + 3

= 2 × x2 + 2 × 3 × x + 1x + 3   [ find the common ]

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

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

(iii) 6x2 + 5x - 6

Solution :

We have 6x2 + 5x - 6

[ Using the splitting the middle term method ]

So, 6x2 + 5x - 6

= 6x2 + 9x - 4x - 6

= 2 × 3 × x2 + 3 × 3 × x - 2 × 2 × x - 2 × 3    [ find the common ]

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

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

(iv) 3x2 - x - 4

Solution :

We have 3x2 - x - 4

[ Using the splitting the middle term method ]

So, 3x2 - x - 4

= 3x2 - 4x + 3x - 4

= 3 × x2 - 2 × 2 × x + 3 × x - 2 × 2   [ find the common ]

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

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

Q5. Factorise :

(i) x3 - 2x2 - x + 2

Solution : 

Let p( x ) = x3 - 2x2 - x + 2

So, x - 1 = 0

x = 1

When p(x) = x3 - 2x2 - x + 2

than, ( 1 )3 - 2( 1 )2 - ( 1 ) + 2

= 1 - 2 × 1 - 1 + 2

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

= 0

∴ ( x - 1 ) is a factor of p( x )

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Q.5, (i)

So, p( x ) = x2 - x - 2

x2 - x - 2

[ Using the splitting the middle term method ]

So, x2 - x - 2

= x2 - 2x + x - 2     [ find the common ]

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

= ( x + 1 ) ( x - 2 )

∴ ( x + 1 ) ( x + 1 ) ( x - 2 ) is a factor of p( x ) .

(ii) x3 - 3x2 - 9x - 5

Solution :

Let p( x ) = x3 - 3x2 - 9x - 5

So, x + 1 = 0

x = - 1

When p(x) = x3 - 2x2 - x + 2

than, ( - 1 )3 - 3( - 1 )2 - 9( - 1 ) - 5

= - 1 - 3 × 1 + 9 - 5

= - 1 - 3 + 9 - 5

= - 9 + 9

= 0

∴ ( x + 1 ) is a factor of p( x )

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Q.5, (ii)

So, p( x ) = x2 - 4x - 5

x2 - 4x - 5

[ Using the splitting the middle term method ]

So, x2 - 4x - 5

= x2 - 5x + x - 5     [ find the common ]

= x ( x - 5 ) + 1 ( x - 5 )

= ( x + 1 ) ( x - 5 )

∴ ( x + 1 ) ( x - 5 ) ( x + 1 ) is a factor of p( x ) .

(iii) x3 + 13x2 + 32x + 20

Solution :

Let p( x ) =  x3 + 13x2 + 32x + 20

So, x + 1 = 0

x = - 1

When p(x) =  x3 + 13x2 + 32x + 20

than, ( - 1 )3 + 13( - 1 )2 + 32( - 1 ) + 20

= - 1 + 13 × 1 - 32 + 20

= - 1 + 13 - 32 + 20

= - 33 + 33

= 0

∴ ( x + 1 ) is a factor of p( x )

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Q.5, (iii)

So, p( x ) = x2 + 12x + 20

x2 + 12x + 20

[ Using the splitting the middle term method ]

So, x2 + 12x + 20

= x2 + 10x + 2x + 20

= x ( x + 10 ) + 2 ( x + 10 )

= ( x + 2 ) ( x + 10 )

∴ ( x + 1 ) ( x + 2 ) ( x + 10 ) is a factor of p( x ) .

(iv) 2y3 + y2 - 2y - 1

Solution :

Let p( y ) = 2y3 + y2 - 2y - 1

So, y + 1 = 0

y = - 1

When p(y) = 2y3 + y2 - 2y - 1

than, 2( - 1 )3 + ( - 1 )2 - 2( - 1 ) - 1

= 2 × - 1 + 1 + 2 - 1

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

= 0

∴ ( y + 1 ) is a factor of p( x )

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.4 Q.5, (iv)

So, p( y ) = 2y2 - y - 1

2y2 - y - 1

[ Using the splitting the middle term method ]

So, 2y2 - y - 1

= 2y2 - 2y + y - 1

= 2y ( y - 1 ) + 1 ( y - 1 )

= ( 2y + 1 ) ( y - 1 )

∴ ( y + 1 ) ( 2y + 1 ) ( y - 1 ) is a factor of p( y ) .

 

NCERT Class 9 Maths Chapter 2 Polynomials :

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