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# Class 9 SELINA Solutions Maths Chapter 5 - Factorisation

## Factorisation Exercise Ex. 5(A)

### Solution 1

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

### Solution 2

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]

### Solution 3

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]

### Solution 4

2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)

= 2x(a - b) + 15y(a - b) - 8z(a - b)

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

### Solution 5

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

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

### Solution 6

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)

### Solution 17

= (x2 + y2 + 2xy ) + (x + y)

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

=(x + y)2 + (x + y)

=(x + y)(x + y + 1)

### Solution 18

= 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)

### Solution 19

= 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)

### Solution 20

=(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)

## Factorisation Exercise Ex. 5(B)

### Solution 17

(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)

### Solution 20

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)

## Factorisation Exercise Ex. 5(D)

### Solution 9

= (x - y)3 - (2x)3

= (x - y - 2x)[(x - y)2 + 2x(x - y) + (2x)2]

[Using identity (a3 - b3) = (a - b)(a2 + ab + b2)]

= (-x - y)[x2 + y2 - 2xy + 2x2 - 2xy + 4x2]

=-(x + y) [7x2 - 4xy + y2]

### Solution 16

= 2(x3 + 27y3 - 2x - 6y)

= 2{[(x)3+(3y)3] - 2(x  + 3y)}

[Using identity (a3 +  b3) = (a + b)(a2 - ab + b2)]

=2{[(x + 3y)(x2 - 3xy + 9y2)] - 2(x + 3y)}

=2(x + 3y)(x2 - 3xy + 9y2 - 2)

### Solution 17

1029 - 3x3

= 3(343 - x3)

= 3(73 - x3)

= 3(7 - x)(72 + 7x + x2)

= 3(7 - x)(49 + 7x + x2)

### Solution 18

(i) (133 - 53)

[Using identity (a3 - b3) = (a - b)(a2 + ab + b2)]

=(13 - 5)(132 + 13 × 5 + 52)

=8(169 + 65 + 25)

Therefore, the number is divisible by 8.

(ii) (353 + 273)

[Using identity (a3 + b3)=(a + b)(a2 - ab + b2)]

=(35 + 27)(352 + 35× 27 + 272)

=62 × (352 + 35 × 27 + 272)

Therefore, the number is divisible by 62.