SELINA Solution Class 9 Factorisation Chapter 5 Exercise 5E

Question 1

Factorise : ${\displaystyle x^2+\frac{1}{{\left(4x\right)}^2}+1-7x-\frac{7}{2x}}$

Sol:

${\displaystyle x^2+\frac{1}{{\left(4x\right)}^2}+1-7x-\frac{7}{2x}}$

= ${\displaystyle {\left(x\right)}^2+\frac{1}{{\left(2x\right)}^2}+2\times x\times \frac{1}{2x}-7(x+\frac{1}{2x})}$

= ${\displaystyle {(x+\frac{1}{2x})}^2-7(x+\frac{1}{2x})}$

= ${\displaystyle (x+\frac{1}{2x})(x+\frac{1}{2x}-7)}$

= ${\displaystyle (x+\frac{1}{2x})(x-7+\frac{1}{2x})}$

Question 2

Factorise : ${\displaystyle {\left(9a\right)}^2+\frac{1}{{\left(9a\right)}^2}-2-12a+\frac{4}{3a}}$

Sol:

${\displaystyle {\left(9a\right)}^2+\frac{1}{{\left(9a\right)}^2}-2-12a+\frac{4}{3a}}$

= ${\displaystyle {\left(3a\right)}^2+\frac{1}{{\left(3a\right)}^2}-2\times 3a\times \frac{1}{3a}-4(3a-\frac{1}{3a})}$

= ${\displaystyle {(3a-\frac{1}{3a})}^2-4(3a-\frac{1}{3a})}$

= ${\displaystyle (3a-\frac{1}{3a})[(3a-\frac{1}{3a})-4]}$

= ${\displaystyle (3a-\frac{1}{3a})(3a-4-\frac{1}{3a})}$

Question 3

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

Sol:

${\displaystyle x^2+\frac{a^2+1}{a}x+1=0}$

∴ ${\displaystyle x^2+ax+\frac{1}{a}x+1=0}$

∴ ${\displaystyle x(x+a)+\frac{1}{a}(x+a)=0}$

∴ ${\displaystyle (x+a)(x+\frac{1}{a})=0}$

Question 4

Factorise : ${\displaystyle x^4+y^4-27x^2{y}^2}$

Sol:

${\displaystyle x^4+y^4-27x^2{y}^2}$

= ${\displaystyle {\left(x^2\right)}^2+{\left(y^2\right)}^2-2x^2{y}^2-25x^2{y}^2}$

= ${\displaystyle {(x^2-y^2)}^2-25x^2{y}^2}$

= ${\displaystyle {(x^2-y^2)}^2-{\left(5xy\right)}^2}$      [∵ a2 - b2 = ( a + b )( a - b )]

= ${\displaystyle [(x^2-y^2)+5xy][(x^2-y^2)-5xy]}$

= ${\displaystyle [x^2+5xy-y^2][x^2-5xy-y^2]}$

Quesiton 5

Factorise : 4x⁴ + 9y⁴ + 11x²y²

Sol:

4x⁴ + 9y⁴ + 11x²y²

= (2x²)² + (3y²)² + 12x²y² - x²y²

= (2x² + 3y²)² - x²y²

= (2x² + 3y²)² - (xy)²

= ( 2x² + 3y² - xy )( 2x² + 3y² + xy)          [ ∵ a² - b² = ( a + b )( a - b )]

Quesiton 6

Factorise : ${\displaystyle x^2+\frac{1}{x^2}-3}$

Sol:

${\displaystyle x^2+\frac{1}{x^2}-3}$

= ${\displaystyle x^2+\frac{1}{x^2}-2\times x\times \frac{1}{x}-1}$

= ${\displaystyle {(x-\frac{1}{x})}^2-1}$

= ${\displaystyle {(x-\frac{1}{x})}^2-{\left(1\right)}^2}$

= ${\displaystyle (x-\frac{1}{x}-1)(x-\frac{1}{x}+1)}$   [ ∵ a² - b² = ( a + b )( a - b )]

Question 7

Factorise : a - b - 4a² + 4b²

Sol:

a - b - 4a² + 4b²  
= ( a - b ) - 4( a² - b² )

= ( a - b ) - 4( a - b )( a + b )   [ ∵ a² - b² = ( a + b )( a - b )]

= ( a - b )[ 1 - 4( a + b )]

= ( a - b )[ 1 - 4a - 4b ]

Question 8

Factorise : (2a - 3)² - 2 (2a - 3) (a - 1) + (a - 1)²

Sol:

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

= [( 2a - 3 ) - ( a - 1 )]²

= [ 2a - 3 - a + 1 ]²

= ( a - 2 )²

Question 9

Factorise : (a² - 3a) (a² - 3a + 7) + 10

Sol:

(a² - 3a) (a² - 3a + 7) + 10

Let us assume , a² - 3a = x
Then, our polynomial becomes,
( a² - 3a )( a² - 3a + 7 ) + 10
= x( x + 7 ) + 10
= x² + 7x + 10
= x² + 5x + 2x + 10
= x( x + 5 ) + 2 ( x + 5 )
= ( x + 5 )( x + 2 )

By resubstituting the value of x,
= (a² - 3a + 5)( a² - 3a + 2 )

Now, a² - 3a + 5 will have no factor as discriminant is -11 that is less than 0.

And,

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

So, factor of given polynomial are,

a² - 3a + 2 = a² - 2a - a + 2

= (a² - 3a + 5)(a - 1)(a - 2)

Question 10

Factorise : (a² - a) (4a² - 4a - 5) - 6

Sol:

Let us assume, a² - a = x
Then the given expression is,
(a² - a) (4a² - 4a - 5) - 6
= x( 4x - 5 ) - 6
= 4x² - 5x - 6
= 4x² - 8x + 3x - 6
= 4x( x - 2 ) + 3( x - 2 )
= ( 4x + 3 )( x - 2 )
= [ 4( a² - a ) + 3 ]( a² - a - 2 )          [ resubstitute the value of x ]
= [ 4a² - 4a + 3 ]( a² - a - 2 )
= [ 4a² - 4a + 3 ]( a² - 2a + a - 2 )
= [ 4a² - 4a + 3 ][ a( a - 2 ) + 1( a - 2 )]
= [ 4a² - 4a + 3 ]( a - 2 )( a + 1 )

Question 11

Factorise : x⁴ + y⁴ - 3x²y²

Sol:

x⁴ + y⁴ - 3x²y² 
= x⁴ + y⁴ - 2x²y² - x²y²

= (x²)² + (y²)² - 2x²y² - x²y²

= ( x² - y² )² - (xy)²

= ( x² - y² - xy )( x² - y² + xy )         [ ∵ a² - b² = ( a + b )( a - b )]

Question 12

Factorise : 5a² - b² - 4ab + 7a - 7b

Sol:

5a² - b² - 4ab + 7a - 7b
= 4a² + a² - b² - 4ab + 7a - 7b
= a² - b² + 4a² - 4ab + 7a - 7b
= ( a² - b² ) + 4a( a - b ) + 7( a - b )
= ( a - b )( a + b ) + 4a( a - b ) + 7( a - b )     [ ∵ a² - b² = ( a + b )( a - b ) ]
= ( a - b )[ ( a + b ) + 4a + 7 ]
= ( a - b )[ ( a + b ) + 4a + 7 ]
= ( a - b )[ 5a + b + 7 ]

Question 13

Factorise : 12(3x - 2y)² - 3x + 2y - 1

Sol:

12(3x - 2y)² - 3x + 2y - 1 = 12( 3x - 2y )² - ( 3x - 2y ) - 1
Let us assume that 3x - 2y = a
Then the given expression is
12(3x - 2y)² - 3x + 2y - 1
= 12a² - 3a - 1
= 12a² - 4a + 3a - 1
= 4a( 3a - 1 ) + 1( 3a - 1 )
= ( 4a + 1 )( 3a - 1 )
= [ 4( 3x - 2y ) + 1 ][ 3( 3x - 2y ) - 1 ]    [ resubstitute the value of a ]
= ( 12x - 8y + 1 )( 9x - 6y - 1)

Question 14

Factorise : 4(2x - 3y)² - 8x+12y - 3

Sol:

 4(2x - 3y)² - 8x+12y - 3
=  4(2x - 3y)² - 4(2x - 3y) - 3
Let us assume that 2x - 3y = a
Then the given expression is
 4(2x - 3y)² - 8x+12y - 3 
= 4a² - 4a - 3
= 4a² - 6a + 2a - 3 
= 2a( 2a - 3 ) + 1( 2a - 3)
= ( 2a - 3 )( 2a + 1 )
= [ 2( 2x - 3y ) - 3 ][ 2( 2x - 3y ) + 1 ]
= ( 4x - 6y - 3 )( 4x - 6y + 1 )

Question 15

Factorise : 3 - 5x + 5y - 12(x - y)²

Sol:

3 - 5x + 5y - 12(x - y)²  = 3 - 5( x - y ) - 12(x - y)² 
Let us assume that x - y = a
Then the given expression is
3 - 5x + 5y - 12(x - y)² 
= 3 - 5a - 12a²
= 3 - 9a + 4a - 12a²
= 3( 1 - 3a ) + 4a( 1 - 3a )
= ( 3 + 4a )( 1 - 3a )                 [ resubstitute the value of a ]
= [ 3 + 4( x - y )][ 1 - 3( x - y )]
= ( 3 + 4x - 4y )( 1 - 3x +  3y )

Question 16

Factorise : 9x ² + 3x - 8y - 64y²

Sol:

9x ² + 3x - 8y - 64y² 
= 9x² - 64y² + 3x - 8y
= [ (3x)² - (8y)² ] + ( 3x - 8y )
= [( 3x + 8y )( 3x - 8y )] + ( 3x - 8y )
= ( 3x - 8y )( 3x + 8y + 1 )

Question 17

Factorise : 2√3x² + x - 5√3

Sol:

 2√3x² + x - 5√3

= 2√3x² + 6x - 5x - 5√3

= 2√3x( x + √3 ) - 5( x + √3 )

= ( 2√3x - 5 )( x + √3 )

Question 18

Factorise : ${\displaystyle \frac{1}{4}{(a+b)}^2-\frac{9}{16}{(2a-b)}^2}$ 

Sol:

${\displaystyle \frac{1}{4}{(a+b)}^2-\frac{9}{16}{(2a-b)}^2}$ 

=${\displaystyle \frac{1}{4}[{(a+b)}^2-\frac{9}{4}{(2a-b)}^2]}$

=${\displaystyle \frac{1}{4}[{(a+b)}^2-\left[\frac{3}{2}{(2a-b)}^2\right]]}$

=${\displaystyle \frac{1}{4}\left[(a+b+\frac{3}{2}(2a-b))(a+b-\frac{3}{2}(2a-b))\right]}$

=${\displaystyle \frac{1}{4}\left[(a+b+3a-\frac{3b}{2})(a+b-3a+\frac{3b}{2})\right]}$

= ${\displaystyle \frac{1}{4}\left[(4a-\frac{b}{2})(\frac{5b}{2}-2a)\right]}$

= ${\displaystyle \frac{1}{4}\left[\left(\frac{8a-b}{2}\right)\left(\frac{5b-4a}{2}\right)\right]}$

= ${\displaystyle \frac{1}{4}\left[\frac{1}{4}(8a-b)(5b-4a)\right]}$

= ${\displaystyle \frac{1}{16}(8a-b)(5b-4a)}$

Question 19

Factorise : 2(ab + cd) - a² - b² + c² + d²

Sol:

2(ab + cd) - a² - b² + c² + d² 
= 2ab + 2cd - a² - b² + c² + d² 
= c² + d² + 2cd - a² - b² + 2ab
= ( c² + d² + 2cd ) - ( a² + b² - 2ab )
= ( c + d )² - ( a - b )²
= ( c + d + a - b )( c + d - a + b )

Question 20.1

Find the value of : ( 987 )² - (13)²

Sol:

( 987 )² - (13)² 
= ( 987 + 13 )( 987 - 13)
= 1000 x 974
= 974000

Question 20.2

Find the value of : ( 67.8 )² - ( 32.2 )²

Sol:

( 67.8 )² - ( 32.2 )² 
= ( 67.8 + 32.2 )( 67.8 - 32.2 )
= 100 x 35.6
= 3560

Question 20.3

Find the value of : ${\displaystyle \frac{{\left(6.7\right)}^2-{\left(3.3\right)}^2}{6.7-3.3}}$

Sol:

${\displaystyle \frac{{\left(6.7\right)}^2-{\left(3.3\right)}^2}{6.7-3.3}}$

= ${\displaystyle \frac{(6.7+3.3)(6.7-3.3)}{(6.7-3.3)}}$

= 10

Question 20.4

Find the value of : ${\displaystyle \frac{{\left(18.5\right)}^2-{\left(6.5\right)}^2}{18.5+6.5}}$

Sol:

${\displaystyle \frac{{\left(18.5\right)}^2-{\left(6.5\right)}^2}{18.5+6.5}}$

= ${\displaystyle \frac{(18.5+6.5)(18.5-6.5)}{(18.5+6.5)}}$

= 12