SELINA Solutions for Class 9 Maths Chapter 5 - Factorisation

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Chapter 5 - 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 7

 

Solution 8

Solution 9

 

 

 

Solution 10

 

 

 

Solution 11

 

 

 

 

Solution 12

 

 

Solution 13

 

Solution 14

 

Solution 15

 

 

Solution 16

 

 

 

 

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)

 

Chapter 5 - Factorisation Exercise Ex. 5(B)

Solution 1

 

 

Solution 2

 

 

Solution 3

 

 

Solution 4

 

 

 

Solution 5

 

 

 

Solution 6

 

 

 

Solution 7

 

 

 

Solution 8

 

 

 

 

Solution 9

 

 

 

Solution 10

 

 

 

Solution 11

 

 

 

Solution 12

 

 

 

Solution 13

 

 

 

 

Solution 14

 

 

 

Solution 15

 

 

 

 

 

 

 

 

 

 

 

 

Solution 16

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 18

Solution 19

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)

Chapter 5 - Factorisation Exercise Ex. 5(C)

Solution 1

 

 

Solution 2

 

Solution 3

 

 

 

Solution 4

 

 

 

Solution 5

 

 

 

Solution 6

 

 

 

Solution 7

 

 

 

Solution 8

 

 

 

 

Solution 9

 

 

Solution 10

 

 

Solution 11

 

 

Solution 12

 

 

 

Solution 13

 

 

 

Solution 14

 

 

 

Solution 15

 

 

 

Solution 16

 

 

 

Solution 17

 

 

 

Solution 18

 

 

 

Solution 19

 

 

 

 

Solution 20

 

 

Solution 21

 

 

Solution 22

 

Solution 23

 

 

Solution 24

Solution 25

Solution 26

Solution 27

 

Solution 28

 

Solution 29

Solution 30

Chapter 5 - Factorisation Exercise Ex. 5(D)

Solution 1

 

 

 

Solution 2

 

 

Solution 3

 

 

 

Solution 4

 

 

 

Solution 5

 

 

Solution 6

 

 

 

Solution 7

 

 

 

Solution 8

 

 

 

 

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 10

fraction numerator 8 straight a cubed over denominator 27 end fraction minus straight b cubed over 8 equals open parentheses fraction numerator 2 straight a over denominator 3 end fraction close parentheses cubed minus open parentheses straight b over 2 close parentheses cubed space space space space space space space space space space space space space space space space space space space space space equals open parentheses fraction numerator 2 straight a over denominator 3 end fraction minus straight b over 2 close parentheses open square brackets open parentheses fraction numerator 2 straight a over denominator 3 end fraction close parentheses squared plus fraction numerator 2 straight a over denominator 3 end fraction cross times straight b over 2 plus open parentheses straight b over 2 close parentheses squared close square brackets left square bracket because straight a cubed space minus space straight b cubed equals left parenthesis straight a minus straight b right parenthesis left parenthesis straight a squared equals ab plus straight b squared right parenthesis right square bracket space space space space space space space space space space space space space space space space space space space space space equals open parentheses fraction numerator 2 straight a over denominator 3 end fraction minus straight b over 2 close parentheses open square brackets fraction numerator 4 straight a squared over denominator 9 end fraction plus ab over 3 plus straight b squared over 4 close square brackets

Solution 11

 

 

 

Solution 12

 

 

Solution 13

 

 

Solution 14

 

 

Solution 15

 

 

 

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.

 

Solution 19

Chapter 5 - Factorisation Exercise Ex. 5(E)

Solution 1

 

 

 

 

 

 

Solution 2

 

 

 

 

 

 

 

 

 

Solution 3

 

 

 

 

Solution 4

 

 

 

Solution 5

 

 

 

 

 

Solution 6

 

 

 

 

Solution 7

 

 

 

 

 

 

 

Solution 8

 

 

 

 

 

Solution 9

 

 

Solution 10

 

 

 

 

Solution 11

 

 

 

 

Solution 12

Solution 13

 

 

 

 

 

Solution 14

 

 

 

 

Solution 15

 

 

 

 

Solution 16

Solution 17

 

Solution 18

 

Solution 19

Solution 20