SELINA Solution Class 9 Factorisation Chapter 5 Exercise 5B

Question 1

Factorise : a² + 10a + 24

Sol:

a² + 10a + 24
= a² + 6a + 4a + 24
= a( a + 6 ) + 4( a + 6 )
= ( a + 6 )( a + 4 )

Question 2

Factorise : a² - 3a - 40

Sol:

a² - 3a - 40
= a² - 8a + 5a - 40
= a( a - 8 ) + 5( a - 8 )
= ( a - 8 )( a + 5 )

Question 3

Factorise : 1 - 2a - 3a²

Sol:

1 - 2a - 3a² 
= 1 - 3a + a - 3a2
=( 1 + a )( 1 - 3a )

Question 4

Factorise : x² - 3ax - 88a²

Sol:

x² - 3ax - 88a²
= x² - 11ax + 8ax - 88a²
= x( x - 11a ) + 8a( x - 11a )
= ( x + 8a )( x - 11a )

Question 5

Factorise : 6a² - a - 15

Sol:

6a² - a - 15
= 6a² - 10a + 9a - 15
= 2a( 3a - 5 ) + 3( 3a - 5 )
= ( 2a + 3 )( 3a - 5 )

Question 6

Factorise : 24a³ + 37a² - 5a

Sol:

24a³ + 37a² - 5a 
= a( 24a2 + 37a - 5 )
= a( 24a2 + 40a -3a - 5 )
= a x [ 8a( 3a + 5 ) - 1( 3a + 5 )]
= a[( 8a - 1 )( 3a + 5 )]
= a( 8a - 1 )( 3a + 5 )

Question 7

Factorise : a(3a - 2) - 1

Sol:

a(3a - 2) - 1
= 3a2 - 2a - 1
= 3a2 - 3a + a - 1
= 3a( a - 1 ) + 1( a - 1 )
= ( 3a + 1 )( a - 1 )

Question 8

Factorise : a²b² + 8ab - 9

Sol:

 a²b² + 8ab - 9 
=  a²b² + 9ab - ab - 9
= ab( ab + 9 ) -1( ab + 9 )
= ( ab + 9 )( ab - 1 )

Question 9

Factorise : 3 - a (4 + 7a)

Sol:

3 - a (4 + 7a) 
= 3 - 4a - 7a2
= 3 - 7a + 3a - 7a2
= 1( 3 - 7a ) + a( 3 - 7a )
= ( 3 - 7a )( a + 1 )

Question 10

Factorise : (2a + b)² - 6a - 3b - 4

Sol:

(2a + b)² - 6a - 3b - 4
= ( 2a + b )² - 3( 2a + b ) - 4
Assume that, 2a + b = x
Therefore,
(2a + b)² - 6a - 3b - 4
= x2 - 3x - 4 
= x2 - 4x + x - 4
= 1( x - 4 ) + x( x - 4 )
= ( x + 1 )( x - 4 )
= ( 2a + b + 1 )( 2a + b - 4 ) 
[ resubstitute the value of x ]

Question 11

Factorise : 1 - 2 (a+ b) - 3 (a + b)²

Sol:

1 - 2 (a+ b) - 3 (a + b)²
Assume that a + b = x ;
1 - 2( a + b ) - 3( a + b )2 
= 1 - 2x - 3x2
= 1 - 3x + x - 3x2
= 1( 1 - 3x ) + x( 1 - 3x )
= ( 1 - 3x )( 1 + x )
= [ 1 - 3( a + b )][ 1 + ( a + b )]
= ( 1 - 3a - 3b )( 1 + a + b )

Question 12

Factorise : 3a² - 1 - 2a

Sol:

3a² - 1 - 2a 
= 3a2 - 2a - 1
= 3a2 - 3a + a - 1
= 3a( a - 1 ) + 1( a - 1 )
= ( 3a + 1 )( a - 1 )

Question 13

Factorise : x² + 3x + 2 + ax + 2a

Sol:

x² + 3x + 2 + ax + 2a 
= x² + 2x + x + 2 + ax + 2a
= x( x + 2 )+1( x + 2 ) +a( x + 2 )
= ( x + 2 )( x + a + 1 )

Question 14

Factorise : (3x - 2y)² + 3 (3x - 2y) - 10

Sol:

(3x - 2y)² + 3 (3x - 2y) - 10
Assume that 3x - 2y = a
Therefore,
(3x - 2y)² + 3 (3x - 2y) - 10
= a2 + 3a - 10
= a2 + 5a - 2a -10
= a( a + 5 ) -2 ( a + 5 )
= ( a + 5 )( a - 2 )
= ( 3x - 2y + 5 )( 3x - 2y - 2)

Question 15

Factorise : 5 - (3a² - 2a) (6 - 3a² + 2a)

Sol:

5 - (3a² - 2a) (6 - 3a² + 2a)
= 5 - ( 3a² - 2a )[ 6 - ( 3a² - 2a )]
Assume that 3a² - 2a = x
Therefore,
5 - ( 3a² - 2a )( 6 - 3a² + 2a )
= 5 - x( 6 - x )
= 5 - 6x + x²
= 5( 1 - x ) - x( 1 - x )
= ( 5 - x )( 1 - x )
= ( x - 5 )( x - 1 )
= ( 3a² - 2a - 5 )( 3a² - 2a - 1 )
= ( 3a² - 5a + 3a -5 )(3a² - 3a + a - 1 )
= [ a( 3a - 5 ) + 1( 3a - 5)][3a( a - 1) + 1( a - 1)]
= ( 3a - 5 )( a + 1 )( 3a + 1 )( a - 1 )

Question 16

Factorise : ${\displaystyle \frac{1}{35}+\frac{12}{35}a+a^2}$

Sol:

${\displaystyle \frac{1}{35}+\frac{12}{35}a+a^2}$

= ${\displaystyle a^2+\frac{12}{35}a+\frac{1}{35}}$

= ${\displaystyle \frac{1}{35}+\frac{12}{35}a+a^2}$

= ${\displaystyle \frac{1}{35}(35a^2+12a+1)}$

= ${\displaystyle \frac{1}{35}(35a^2+12a+1)}$

= ${\displaystyle \frac{1}{35}(35a^2+7a+5a+1)}$

= ${\displaystyle \frac{1}{35}(35a^2+7a+5a+1)}$

= ${\displaystyle \frac{1}{35}(7a(5a+1)+5a+1)}$

= ${\displaystyle \frac{1}{35}(7a(5a+1)+5a+1)}$

= ${\displaystyle \frac{1}{35}(7a+1)(5a+1)}$

Question 17

Factories: (x² - 3x)(x² - 3x - 1) - 20.

Sol:

(x² - 3x)(x² - 3x - 1) - 20
= (x² - 3x)[(x² - 3x) - 1] - 20
= a[a - 1] - 20                        ….( Taking x² - 3x = a )
= a² - a - 20
= a² - 5a + 4a - 20
= a(a - 5) + 4(a - 5)
= (a - 5)(a + 4)
= (x² - 3x - 5)(x² - 3x + 4)

Question 18.1

Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
x² - 3x - 54

Sol:

Given expression : x² - 3x - 54
Comparing with ax² + bx + c, we get a = 1, b = -3, and c = - 54
∴ b² - 4ac = (-3)² - 4(1)(-54) = 9 + 216 = 225, which is a perfect square.
∴ x² - 3x - 54 is factorisable.
Now, x² - 3x - 54 = x² - 9x + 6x - 54
= x( x - 9 ) + 6( x - 9 )
= ( x - 9 )( x + 6 )

Question 18.2

Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
2x² - 7x - 15

Sol:

Given expression : 2x² - 7x - 15
Comparing with ax² + bx + c, we get a = 2, b = -7, and c = -15
∴ b² - 4ac = (-7)² - 4(2)(-15) = 49 + 120 = 169, which is a perfect square.
∴ 2x² - 7x - 15 is factorisable.
Now, 2x² - 7x - 15
= 2x2 - 10x + 3x - 15
= 2x( x - 5 ) + 3( x - 5 ) 
= ( 2x + 3 )( x - 5 )

Question 18.3

Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
2x² + 2x - 75

Sol:

Given expression : 2x² + 2x - 75
Comparing with ax2 + bx + c, we get a = 2, b = 2, and c = - 75
∴ b2 - 4ac = (2)² - 4(2)(-75) = 4 + 600 = 604, which is not a perfect square.
∴ 2x² + 2x - 75 is not factorisable.

Question 18.4

Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
3x² + 4x - 10

Sol:

Given expression : 3x² + 4x - 10
Comparing with ax2 + bx + c, we get a = 3, b = 4, and c = -10
∴ b2 - 4ac = (4)² - 4(3)(-10) = 16 + 120 = 136, which is not a perfect square.
∴ 3x² + 4x - 10 is not factorisable.

Question 18.5

Find trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.
x(2x - 1) - 1 

Sol:

Given expression : x(2x - 1) - 1 
Now , x(2x - 1) - 1 = 2x² - x - 1
Comparing with ax² + bx + c, we get a = 2, b = - 1, and c = - 1
∴ b² - 4ac = (- 1)² - 4(2)(-1) = 1 + 8 = 9, which is a perfect square.
∴ 2x² - x - 1 is factorisable.
Now, 2x² - x - 1 = 2x2 - 2x + x - 1
                          = 2x( x - 1 ) + 1( x - 1 )
                          = ( 2x + 1 )( x - 1 )

Question 19.1

Factorise : 4√3x² + 5x - 2√3

Sol:

4√3x² + 5x - 2√3
= 4√3x² + 8x - 3x - 2√3
= 4x( √3x + 2 ) - √3( √3x + 2 )
= ( √3x + 2 )( 4x - √3 )

Question 19.2

Factorise : 7√2x² - 10x - 4√2

Sol:

7√2x² - 10x - 4√2 
= 7√2x² - 14x + 4x - 4√2 
= 7√2x( x - √2 ) + 4( x - √2 )
= ( x - √2 )( 7√2x + 4 )

Question 20

Give possible expressions for the length and the breadth of the rectangle whose area is 12x² - 35x + 25

Sol:

12x² - 35x + 25
= 12x² - 20x - 15x + 25
= 4x(3x - 5) - 5(3x - 5)
= (3x - 5)(4x - 5)
Thus,
Length = (3x - 5) and breadth = (4x - 5)
OR
Length = (4x - 5) and breadth = (3x - 5)