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 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)
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)
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
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
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
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
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