Chapter 5 : Factorisation - Selina Solutions for Class 9 Maths ICSE

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

Read  more

Chapter 5 - Factorisation Excercise Ex. 5(E)

Question 1

Factorize :

 

Solution 1

 

 

 

 

 

 

Question 2

Factorize :

 

Solution 2

 

 

 

 

 

 

 

 

 

Question 3

Factorize :

 

 

Solution 3

 

 

 

 

Question 4

Factorize :

Solution 4

 

 

 

Question 5

Factorize :

4x4 + 9y4 + 11x2y2

Solution 5

 

 

 

 

 

Question 6

Factorize :

 

Solution 6

 

 

 

 

Question 7

Factorize :

a - b - 4a2 + 4b2

Solution 7

 

 

 

 

 

 

 

Question 8

Factorize :

(2a - 3)2 - 2 (2a - 3) (a - 1) + (a - 1)2

Solution 8

 

 

 

 

 

Question 9

Factorize :

(a2 - 3a) (a2 + 3a + 7) + 10

Solution 9

 

 

Question 10

Factorize :

(a2 - a) (4a2 - 4a - 5) - 6

Solution 10

 

 

 

 

Question 11

Factorize :

x4 + y4 - 3x2y2

Solution 11

 

 

 

 

Question 12

Factorize :

5a2 - b2 - 4ab + 7a - 7b

Solution 12

Question 13

Factorize :

12(3x - 2y)2 - 3x + 2y - 1

Solution 13

 

 

 

 

 

Question 14

Factorize :

4(2x - 3y)2 - 8x+12y - 3

Solution 14

 

 

 

 

Question 15

Factorize :

3 - 5x + 5y - 12(x - y)2

Solution 15

 

 

 

 

Question 16

9x 2 + 3x - 8y - 64y2

Solution 16

Question 17

Solution 17

  

Question 18

Solution 18

  

Question 19

2(ab + cd) - a2 - b2 + c2 + d2

Solution 19

Question 20

Solution 20

  

Chapter 5 - Factorisation Excercise Ex. 5(A)

Question 1

Factorise by taking out the common factors:

 

 

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

 

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

 

Question 2

Factories by taking out common factors:

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

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]

Question 3

Factories by taking out common factors:

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

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]

Question 4

Factories by taking out common factors:

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

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]

Question 5

Factorize by the grouping method:

a3 + a - 3a2 - 3

Solution 5

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

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

Question 6

Factorize by the grouping method:

16 (a + b)2 - 4a - 4b

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)

Question 7

Factorize by the grouping method:

a4 - 2a3 - 4a + 8

Solution 7

 

Question 8

Factorize by the grouping method:

ab - 2b + a2 - 2a

Solution 8

Question 9

Factorize by the grouping method:

ab (x2 + 1) + x (a2 + b2)

Solution 9

 

 

 

Question 10

Factorize by the grouping method:

a2 + b - ab - a

Solution 10

 

 

 

Question 11

Factorize by the grouping method:

(ax + by)2 + (bx - ay)2

Solution 11

 

 

 

 

Question 12

Factorize by the grouping method:

a2x2 + (ax2 + 1) x + a

Solution 12

 

 

Question 13

Factorize by the grouping method:

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

Solution 13

 

Question 14

Factorize by the grouping method:

a (a -4) - a + 4

Solution 14

 

Question 15

Factorize by the grouping method:

y2 - (a + b) y + ab

Solution 15

 

 

Question 16

Factorize by the grouping method:

 

 

Solution 16

 

 

 

 

Question 17

Factorise using the grouping method:

 

 

x2 + y2 + x + y + 2xy

 

Solution 17

= (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

 

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)

 

Question 19

Factorise using the grouping method:

 

 

m (x - 3y)2 + n (3y - x) + 5x - 15y

 

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)

 

Question 20

Factorise using the grouping method:

 

 

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

 

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 Excercise Ex. 5(B)

Question 1

Factorize:

a2 + 10a + 24

Solution 1

 

 

Question 2

Factorize:

a2 - 3a - 40

Solution 2

 

 

Question 3

Factorize:

1 - 2a - 3a2

Solution 3

 

 

Question 4

Factorize:

x2 - 3ax - 88a2

Solution 4

 

 

 

Question 5

Factorize:

6a2 - a-15

Solution 5

 

 

 

Question 6

Factorize:

24a3 + 37a2 - 5a

Solution 6

 

 

 

Question 7

Factorize:

a(3a - 2) - 1

Solution 7

 

 

 

Question 8

Factorize:

a2b2 + 8ab - 9

Solution 8

 

 

 

 

Question 9

Factorize:

3 - a (4 + 7a)

Solution 9

 

 

 

Question 10

Factorize:

(2a + b)2 - 6a - 3b - 4

Solution 10

 

 

 

Question 11

Factorize:

1 - 2 (a+ b) - 3 (a + b)2

Solution 11

 

 

 

Question 12

Factorize:

3a2 - 1 - 2a

Solution 12

 

 

 

Question 13

Factorize:

x2 + 3x + 2 + ax + 2a

Solution 13

 

 

 

 

Question 14

Factorize:

(3x - 2y)2 + 3 (3x - 2y) - 10

Solution 14

 

 

 

Question 15

Factorize:

5 - (3a2 - 2a) (6 - 3a2 + 2a)

Solution 15

 

 

 

 

 

 

 

 

 

 

 

 

Question 16

Solution 16

Question 17

Factories: (x2 - 3x)(x2 - 3x - 1) - 20.

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)

Question 18

Find each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.

(i) x2 - 3x - 54

(ii) 2x2 - 7x - 15

(iii) 2x2 + 2x - 75

(iv) 3x2 + 4x - 10

(v) x(2x - 1) - 1 

Solution 18

Question 19

Solution 19

Question 20

Give possible expressions for the length and the breadth of the rectangle whose area is

12x2 - 35x + 25

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 Excercise Ex. 5(C)

Question 1

Factorize :

25a2 - 9b2

Solution 1

 

 

Question 2

Factorize :

a2 - (2a + 3b)2

Solution 2

 

Question 3

Factorize :

a2 - 81 (b-c)2

Solution 3

 

 

 

Question 4

Factorize :

25(2a - b)2 - 81b2

Solution 4

 

 

 

Question 5

Factorize :

50a3 - 2a

Solution 5

 

 

 

Question 6

Factorize :

4a2b - 9b3

Solution 6

 

 

 

Question 7

Factorize :

3a5 - 108a3

Solution 7

 

 

 

Question 8

Factorize :

9(a - 2)2 - 16(a + 2)2

Solution 8

 

 

 

 

Question 9

Factorize :

a4 - 1

Solution 9

 

 

Question 10

Factorize :

a3 + 2a2 - a-2

Solution 10

 

 

Question 11

Factorize :

(a + b)3 - a - b

Solution 11

 

 

Question 12

Factorize :

a (a - 1) - b (b - 1)

Solution 12

 

 

 

Question 13

Factorize :

4a2 - (4b2 + 4bc + c2)

Solution 13

 

 

 

Question 14

Factorize :

4a2 - 49b2 + 2a - 7b

Solution 14

 

 

 

Question 15

Factorize :

9a2 + 3a - 8b - 64b2

Solution 15

 

 

 

Question 16

Factorize :

4a2 - 12a + 9 - 49b2

Solution 16

 

 

 

Question 17

Factorize :

4xy - x2 - 4y2 + z2

Solution 17

 

 

 

Question 18

Factorize :

a2 + b2 - c2 - d2 + 2ab - 2cd

Solution 18

 

 

 

Question 19

Factorize :

4x2 - 12ax - y2 - z2 - 2yz + 9a2

Solution 19

 

 

 

 

Question 20

Factorize :

(a2 - 1) (b2 - 1) + 4ab

Solution 20

 

 

Question 21

Factorize :

x4 + x2 + 1

Solution 21

 

 

Question 22

Factorize :

(a2 + b2 - 4c2)2 - 4a2b2

Solution 22

 

Question 23

Factorize :

(x2 + 4y2 - 9z2)2 - 16x2y2

Solution 23

 

 

Question 24

(a + b) 2 - a2 + b2

Solution 24

Question 25

a2 - b2 - (a + b) 2

Solution 25

Question 26

9a2 - (a2 - 4) 2

Solution 26

Question 27

Solution 27

  

Question 28

Solution 28

  

Question 29

4x4 - x2 - 12x - 36

Solution 29

Question 30

a2 ( b + c) - (b + c)3

Solution 30

Chapter 5 - Factorisation Excercise Ex. 5(D)

Question 1

Factorize :

a3 - 27

Solution 1

 

 

 

Question 2

Factorize :

1 - 8a3

Solution 2

 

 

Question 3

Factorize :

64 - a3b3

Solution 3

 

 

 

Question 4

Factorize :

a6 + 27b3

Solution 4

 

 

 

Question 5

Factorize :

3x7y - 81x4y4

Solution 5

 

 

Question 6

Factorize :

a3 -

Solution 6

 

 

 

Question 7

Factorize :

a3 + 0.064

Solution 7

 

 

 

Question 8

Factorize :

a4 - 343a

Solution 8

 

 

 

 

Question 9

Factorise:

 

 

(x - y)3 - 8x3

 

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]

 

Question 10

Factorize :

 

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

Question 11

Factorize :

a6 - b6

Solution 11

 

 

 

Question 12

Factorize :

a6 - 7a3 - 8

Solution 12

 

 

Question 13

Factorize :

a3 - 27b3 + 2a2b - 6ab2

Solution 13

 

 

Question 14

Factorize :

8a3 - b3 - 4ax + 2bx

Solution 14

 

 

Question 15

Factorize :

a - b - a3 + b3

Solution 15

 

 

 

Question 16

Factorise:

 

 

2x3 + 54y3 - 4x - 12y

 

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)

 

Question 17

1029 - 3x3

Solution 17

1029 - 3x3

= 3(343 - x3)

= 3(73 - x3)

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

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

Question 18

Show that:

 

 

(i) 133 - 53 is divisible by 8

 

 

(ii)353 + 273 is divisible by 62

 

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.

 

Question 19

Solution 19

Key Features of Study Materials for ICSE Class 9 Maths:

  • Include video lessons, question bank and solved sample papers
  • Designed according to the latest ICSE syllabus
  • Developed by subject experts
  • Content revised from time to time
  • Helpful when doing quick revision
  • Guidance when doing homework
  • Understand concepts easily
  • Significantly improve your Mathematics score