SELINA Solutions for Class 9 Maths Chapter 5 - Factorisation
Revise the different methods of factorisation with TopperLearning’s Selina Solutions for ICSE Class 9 Mathematics Chapter 5 Factorisation online. While trying to solve factorisation problems, explore our chapter solutions and learn the application of methods like splitting of the middle terms, grouping and more.
You can also revise problems based on factorisation by taking out the common factors when you practise with our Selina textbook solutions for ICSE Class 9 Maths. You may also use our Frank solutions to practise more problems - ensuring you have sufficient revised and feel confident. Also, browse through our other e-learning resources like practice tests, sample papers and more.
Chapter 5 - Factorisation Exercise 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 Exercise 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 Exercise 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 Exercise 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 Exercise 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
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
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