Factorisation ICSE Class-9th Concise Mathematics Selina Solutions Chapter-5. We provide step by step Solutions of Exercise / lesson-5 Factorisation for ICSE Class-9 Concise Selina Mathematics by RK Bansal.

Our Solutions contain all type Questions with Exe-5 A, Exe-5 B, Exe-5 C, Exe-5 D and Exe-5 E to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

## Factorisation ICSE Class-9th Concise Mathematics Selina Solutions Chapter-5

–: Select Topics :–

Exe-5 A,

Exe-5 B,

Exe-5 C,

Exe-5 D,

Exe-5 E,

### Selina Solutions Exercise – 5(A),Factorisation for ICSE Class-9th Concise Mathematics

#### Question 1

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

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(3x– 2y2) – yz(2y– 3x2) + zx(15x– 10y2)

xy(3x– 2y2) – yz(2y– 3x2) + zx(15x– 10y2)

= xy(3x– 2y2) + yz(3x– 2y2) + zx(15x– 10y2)

= xy(3x– 2y2) + yz(3x– 2y2) + 5zx(3x– 2y2)

= (3x2 – 2y2)[xy + yz + 5zx]

#### Question 3

Factories by taking out common factors :
ab(a+ b– c2) – bc(c– a– b2) + ca(a+ b– c2)

ab(a+ b– c2) – bc(c– a– b2) + ca(a+ b– c2)

= ab(a+ b– c2) + bc(a+ b– c2) + ca(a+ b– c2)

= (a+ b– c2)[ab + bc + ca]

#### Question 4

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

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

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

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

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

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)

ab (x2 + 1) + x (a2 + b2) = abx2 + ab + a2x + b2x
= ax( bx + a ) + b( bx + a )
= ( ax + b )( bx + a )

#### Question 10

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

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

(ax + by)2 + (bx – ay)
= 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

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

( 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

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

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 : #### Question 17

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

x2 + y2 + x + y + 2xy

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

#### Question 18

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

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)– 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

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

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)

### Exe-5 B , Factorisation ICSE Class-9th Concise Mathematics Selina Solutions

#### Question 1

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