SELINA Solution Class 9 Factorisation Chapter5 Exercise 5A

Question 1

Factorise by taking out the common factors : 
2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)

Sol:

2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)

Taking (2x - 5y) common from both terms
= (2x - 5y)[2(3x + 4y) - 6(x - y)]
= (2x - 5y)(6x + 8y - 6x + 6y)
= (2x - 5y)(8y + 6y)
= (2x - 5y)(14y)
= (2x - 5y)14y

Question 2

Factories by taking out common factors :
xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²)

Sol:

xy(3x² - 2y²) - yz(2y² - 3x²) + zx(15x² - 10y²)

= xy(3x² - 2y²) + yz(3x² - 2y²) + zx(15x² - 10y²)

= xy(3x² - 2y²) + yz(3x² - 2y²) + 5zx(3x² - 2y²)

= (3x² - 2y²)[xy + yz + 5zx]

Question 3

Factories by taking out common factors :
ab(a² + b² - c²) - bc(c² - a² - b²) + ca(a² + b² - c²)

Sol:

ab(a² + b² - c²) - bc(c² - a² - b²) + ca(a² + b² - c²)

= ab(a² + b² - c²) + bc(a² + b² - c²) + ca(a² + b² - c²)

= (a² + b² - c²)[ab + bc + ca]

Question 4

Factories by taking out common factors :
2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)

Sol:

2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)
= 2x(a - b) + 15y(a - b) - 8z(a - b)
= (a - b)[2x + 15y - 8z]

Question 5

Factorise by the grouping method : a³ + a - 3a² - 3

Sol:

a³ + a - 3a² - 3
= a (a² + 1) - 3(a² + 1) 
= (a² + 1) (a -3).

Question 6

Factorise by the grouping method: 16 (a + b)² - 4a - 4b

Sol:

16 (a + b)² - 4a - 4b =16 (a + b)² - 4 (a + b)
= 4 (a + b) [4 (a + b) - 1]
= 4 (a + b) (4a + 4b - 1)

Question 7

Factorise by the grouping method : a⁴ - 2a³ - 4a + 8

Sol:

a⁴ - 2a³ - 4a + 8 = a³( a - 2 ) - 4( a - 2 )
                           = ( a³ - 4 )( a - 2 )

Question 8

Factorise by the grouping method : ab - 2b + a² - 2a

Sol:

ab - 2b + a² - 2a = b( a - 2 ) + a( a - 2 ) 
                            = ( a + b )( a - 2 )

Question 9

Factorise by the grouping method : ab (x² + 1) + x (a² + b²)

Sol:

ab (x² + 1) + x (a² + b²)

= abx² + ab + a²x + b²x

= abx² + b² + a²x + ab                                       

= bx( ax + b) + a( ax + b )

= ( ax + b )( bx + a ).

Question 10

Factorise by the grouping method : a² + b - ab - a

Sol:

a² + b - ab - a = a2 - a + b - ab
                        = a( a - 1) + b( 1 - a )
                        = a(a - 1) - b(a - 1)
                        = (a -1)(a - b)

Question 11

Factorise by the grouping method : (ax + by)² + (bx - ay)²

Sol:

(ax + by)² + (bx - ay)² 
= a²x² + b²y² + 2axby + b²x² + a²y² - 2bxay
= a²x² + b²y² + b²x² + a²y²
= x²( a² + b² ) + y²( a² + b² )
= ( x² + y² )( a² + b² )

Question 12

Factorise by the grouping method : a²x² + (ax² + 1) x + a

Sol;

a²x² + (ax² + 1) x + a 
= a²x² + a + (ax² + 1) x
= a( ax² + 1) + x( ax² + 1)
= ( a + x )( ax² + 1 )

Question 13

Factorise by the grouping method : (2a-b)² -10a + 5b

Sol:

( 2a - b)² - 10a + 5b 
= ( 2a - b )² - 5( 2a - b )
= ( 2a - b )( 2a - b - 5 )

Question 14

Factorise by the grouping method : a (a -4) - a + 4

Sol:

a (a -4) - a + 4 
= a( a - 4 ) -1( a - 4 )
= ( a - 4 )( a - 1 )

Question 15

Factorise by the grouping method : y² - (a + b) y + ab

Sol:

 y² - (a + b) y + ab
= y2 - ay - by + ab
= y( y - a ) - b( y - a )
= ( y - a )( y - b ) 

Question 16

Factorise by the grouping method : 
${\displaystyle a^2+\frac{1}{a^2}-2-3a+\frac{3}{a}}$

Sol:

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

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

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

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

Question 17

Factorise using the grouping method:
x² + y² + x + y + 2xy

Sol:

x² + y² + x + y + 2xy

= ( x² + y² + 2xy ) + ( x + y )     [As (x + y)² = x² + 2xy + y²]
= ( x + y )² + ( x + y )
= ( x + y )( x + y + 1 )

Question 18

Factorise using the grouping method :
a² + 4b² - 3a + 6b - 4ab

Sol:

a² + 4b² - 3a + 6b - 4ab
= a² + 4b² - 4ab - 3a + 6b
= a² + (2b)² - 2 × a × (2b) - 3(a - 2b)    [As (a - b)² = a² - 2ab + b² ]
= (a - 2b)² - 3(a - 2b)
= (a - 2b)[(a - 2b)- 3]
= (a - 2b)(a - 2b - 3)

Question 19

Factorise using the grouping method :
m (x - 3y)² + n (3y - x) + 5x - 15y

Sol:

m (x - 3y)² + n (3y - x) + 5x - 15y
= m (x - 3y)² - n (x - 3y) + 5(x - 3y)     
[Taking (x - 3y) common from all the three terms]
= (x - 3y) [m(x - 3y) - n + 5]
= (x - 3y)(mx - 3my - n + 5)

Question 20

Factorise using the grouping method :
x (6x - 5y) - 4 (6x - 5y)²

Sol:

x (6x - 5y) - 4 (6x - 5y)²
= (6x - 5y)[x - 4(6x - 5y)]
[Taking (6x - 5y) common from the three terms]
= (6x - 5y)(x - 24x + 20y)
= (6x - 5y)(-23x + 20y)
= (6x - 5y)(20y - 23x)