# 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(3x^{2 }- 2y^{2}) - yz(2y^{2 }- 3x^{2}) + zx(15x^{2 }- 10y^{2})

= xy(3x^{2 }- 2y^{2}) + yz(3x^{2 }- 2y^{2}) + zx(15x^{2 }- 10y^{2})

= xy(3x^{2 }- 2y^{2}) + yz(3x^{2 }- 2y^{2}) + 5zx(3x^{2 }- 2y^{2})

= (3x^{2} - 2y^{2})[xy + yz + 5zx]

### Solution 3

ab(a^{2 }+ b^{2 }- c^{2}) - bc(c^{2 }- a^{2 }- b^{2}) + ca(a^{2 }+ b^{2 }- c^{2})

= ab(a^{2 }+ b^{2 }- c^{2}) + bc(a^{2 }+ b^{2 }- c^{2}) + ca(a^{2 }+ b^{2 }- c^{2})

= (a^{2 }+ b^{2 }- c^{2})[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

a^{3} + a - 3a^{2} - 3= a (a^{2} + 1) - 3(a^{2} + 1)

= (a^{2} + 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)

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### Solution 17

= (x^{2} + y^{2} + 2xy ) + (x + y)

[As (x + y)^{2} = x^{2 }+ 2xy + y^{2}]

=(x + y)^{2 }+ (x + y)

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

### Solution 18

= a^{2} + 4b^{2} - 4ab - 3a + 6b

= a^{2} + (2b)^{2} - 2 × a × (2b) - 3(a - 2b)

[As (a - b)^{2} = a^{2} - 2ab + b^{2} ]

=(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)

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### Solution 17

(x^{2 }- 3x)(x^{2 }- 3x - 1) - 20

= (x^{2} - 3x)[(x^{2} - 3x) - 1] - 20

= a[a - 1] - 20 ….(Taking x^{2} - 3x = a)

= a^{2} - a - 20

= a^{2} - 5a + 4a - 20

= a(a - 5) + 4(a - 5)

= (a - 5)(a + 4)

= (x^{2} - 3x - 5)(x^{2} - 3x + 4)

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### Solution 20

12x^{2 }- 35x + 25

= 12x^{2} - 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(C)

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## Factorisation Exercise Ex. 5(D)

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### Solution 9

= (x - y)^{3} - (2x)^{3}

= (x - y - 2x)[(x - y)^{2} + 2x(x - y) + (2x)^{2}]

[Using identity (a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2})]

= (-x - y)[x^{2} + y^{2} - 2xy + 2x^{2} - 2xy + 4x^{2}]

=-(x + y) [7x^{2 }- 4xy + y^{2}]

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### Solution 16

= 2(x^{3} + 27y^{3} - 2x - 6y)

= 2{[(x)^{3}+(3y)^{3}] - 2(x + 3y)}

[Using identity (a^{3} + b^{3}) = (a + b)(a^{2} - ab + b^{2})]

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

=2(x + 3y)(x^{2} - 3xy + 9y^{2 }- 2)

### Solution 17

1029 - 3x^{3}

= 3(343 - x^{3})

= 3(7^{3} - x^{3})

= 3(7 - x)(7^{2} + 7x + x^{2})

= 3(7 - x)(49 + 7x + x^{2})

### Solution 18

(i) (13^{3} - 5^{3})

[Using identity (a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2})]

=(13 - 5)(13^{2 }+ 13 × 5 + 5^{2})

=8(169 + 65 + 25)

Therefore, the number is divisible by 8.

(ii) (35^{3} + 27^{3})

[Using identity (a^{3} + b^{3})=(a + b)(a^{2} - ab + b^{2})]

=(35 + 27)(35^{2} + 35× 27 + 27^{2})

=62 × (35^{2} + 35 × 27 + 27^{2})

Therefore, the number is divisible by 62.

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## Factorisation Exercise Ex. 5(E)

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