# 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 :

- 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 2.4 (Page-43)**

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

**(i) **x^{3} + x^{2} + x + 1

**Solution :**

Let p(x) = x^{3} + x^{2} + x + 1

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

So, x + 1 = 0

x = - 1

Using the factor theorem,

When p(x) = x^{3} + x^{2} + 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 x^{3} + x^{2} + x + 1 .

**(ii) **x^{4} + x^{3} + x^{2} + x + 1

**Solution :**

Let p(x) = x^{4} + x^{3} + x^{2} + x + 1

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

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = x^{4} + x^{3} + x^{2} + 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 x^{4} + x^{3} + x^{2} + x + 1 .

**(iii) **x^{4} + 3x^{3} + 3x^{2} + x + 1

**Solution :**

Let p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

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

So, x + 1 = 0

x = - 1

Using the factor theorem,

when p(x) = x^{4} + 3x^{3} + 3x^{2} + 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 x^{4} + 3x^{3} + 3x^{2} + x + 1 .

**(iv) **x^{3} - x^{2} - ( 2 + √2 ) x + √2

**Solution :**

Let p(x) = x^{3} - x^{2} - ( 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) = x^{3} - x^{2} - ( 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 x^{3} - x^{2} - ( 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 ) = 2x^{3} + x^{2} - 2x - 1 , g( x ) = x + 1

**Solution :**

Given p(x) = 2x^{3} + x^{2} - 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 ) = x^{3} + 3x^{2} + 3x + 1 , g( x ) = x + 2

**Solution :**

Given p(x) = x^{3} + 3x^{2} + 3x + 1

So, x + 2 = 0

x = - 2

Using the factor theorem,

when p(x) = x^{3} + 3x^{2} + 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 ) = x^{3} - 4x^{2} + x + 6 , g( x ) = x - 3

**Solution :**

Given p(x) = x^{3} - 4x^{2} + x + 6

So, x - 3 = 0

x = 3

Using the factor theorem,

when p(x) = x^{3} - 4x^{2} + 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 ) = x^{2} + x + k

**Solution :**

We have p(x) = x^{2} + x + k

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p( x ) = x^{2} + 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 ) = 2x^{2} + kx + √2

**Solution :**

We have p(x) = 2x^{2} + kx + √2

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = 2x^{2} + 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 ) = kx^{2} - √2x + 1

**Solution :**

We have p(x) = kx^{2} - √2x + 1

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = kx^{2} - √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 ) = kx^{2} - 3x + k

**Solution :**

We have p(x) = kx^{2} - 3x + k

Given x - 1 is factor of p(x)

So, x - 1 = 0

x = 1

Using the factor theorem,

when p(x) = kx^{2} - 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) **12x^{2} - 7x + 1

**Solution :**

We have 12x^{2} - 7x + 1

[ Using the splitting the middle term method ]

So, 12x^{2} - 7x + 1

= 12x^{2} - 4x + 3x + 1

= 3 × 4 × x^{2} - 4x - 3x + 1 [ find the common ]

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

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

**(ii) **2x^{2} + 7x + 3

**Solution :**

We have 2x^{2} + 7x + 3

[ Using the splitting the middle term method ]

So, 2x^{2} + 7x + 3

= 2x^{2} + 6x + 1x + 3

= 2 × x^{2} + 2 × 3 × x + 1x + 3 [ find the common ]

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

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

**(iii) **6x^{2} + 5x - 6

**Solution :**

We have 6x^{2} + 5x - 6

[ Using the splitting the middle term method ]

So, 6x^{2} + 5x - 6

= 6x^{2} + 9x - 4x - 6

= 2 × 3 × x^{2} + 3 × 3 × x - 2 × 2 × x - 2 × 3 [ find the common ]

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

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

**(iv) **3x^{2} - x - 4

**Solution :**

We have 3x^{2} - x - 4

[ Using the splitting the middle term method ]

So, 3x^{2} - x - 4

= 3x^{2} - 4x + 3x - 4

= 3 × x^{2} - 2 × 2 × x + 3 × x - 2 × 2 [ find the common ]

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

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

**Q5. **Factorise :

**(i) **x^{3} - 2x^{2} - x + 2

**Solution : **

Let p( x ) = x^{3} - 2x^{2} - x + 2

So, x - 1 = 0

x = 1

When p(x) = x^{3} - 2x^{2} - 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 )

So, p( x ) = x^{2} - x - 2

x^{2} - x - 2

[ Using the splitting the middle term method ]

So, x^{2} - x - 2

= x^{2} - 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) **x^{3} - 3x^{2} - 9x - 5

**Solution :**

Let p( x ) = x^{3} - 3x^{2} - 9x - 5

So, x + 1 = 0

x = - 1

When p(x) = x^{3} - 2x^{2} - 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 )

So, p( x ) = x^{2} - 4x - 5

x^{2} - 4x - 5

[ Using the splitting the middle term method ]

So, x^{2} - 4x - 5

= x^{2} - 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) **x^{3} + 13x^{2} + 32x + 20

**Solution :**

Let p( x ) = ** **x^{3} + 13x^{2} + 32x + 20

So, x + 1 = 0

x = - 1

When p(x) = ** **x^{3} + 13x^{2} + 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 )

So, p( x ) = x^{2} + 12x + 20

x^{2} + 12x + 20

[ Using the splitting the middle term method ]

So, x^{2} + 12x + 20

= x^{2} + 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) **2y^{3} + y^{2} - 2y - 1

**Solution :**

Let p( y ) = 2y^{3} + y^{2} - 2y - 1

So, y + 1 = 0

y = - 1

When p(y) = 2y^{3} + y^{2} - 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 )

So, p( y ) = 2y^{2} - y - 1

2y^{2} - y - 1

[ Using the splitting the middle term method ]

So, 2y^{2} - y - 1

= 2y^{2} - 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 ) .