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Q.
Factorise the following expressions.
(i) 7x – 42 (ii) 6p – 12q
(iii) 7a2 + 14a (iv) – 16 z + 20 z3
(v) 20 l2 m + 30 a l m (vi) 5 x2 y – 15 xy2
(vii) 10 a 2 – 15 b2 + 20 c2 (viii) – 4 a2 + 4 ab – 4 ca
(ix) x 2 y z + x y2 z + x y z2 (x) a x2 y + b x y2 + c x y z
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Detailed Solution
We know that, Factorization means finding out the factors of the given algebraic expression
(i) 7x 42 :
let 7x, 42 can be written as
7x = 7 x
42 = 2 3 7
The common factors is given as 7
Hence, we get Algebraic expression as
7x 42 = (7 x) (2 3 7 )
= 7 (x 6)
(ii) 6p – 12q :
let 6p, 12q can be written as
6p = 2 3 p
12q = 2 2 3 q
The common factors are given as 2,3
Hence, we get Algebraic expression as
6p – 12q = ( 2 3 p) ( 2 2 3 q )
= 2 3 (p 2 q)
= 6 (p 2q)
(iii) 7a2 14a :
let 7a2, 14a can be written as
7a2 = 7 a a
14a = 2 7 a
The common factors are given as 2, a
Hence, we get Algebraic expression as
7a2 14a = ( 7 a a) ( 2 7 a)
= 7 a (a 2 )
= 7a (a 2)
(iv) – 16 z + 20 z3 :
let – 16 z , 20 z3 can be written as
– 16 z = -1 2 2 2 2 z
20 z3= 2 2 5 z z z
The common factors are given as 2, 2, a
Hence, we get Algebraic expression as
– 16 z 20 z3= ( -1 2 2 2 2 z) (2 2 5 z z z)
= 2 2 z (- 4z 5z2 )
= 4z(- 4z 5z2)
(v) 20 l2 m + 30 a l m :
let 20 l2 m, 30 a l m can be written as
20 l2 m = 2 2 5 I I m
30 a l m = 2 3 5 a l m
The common factors are given as 2, 5, I, m
Hence, we get Algebraic expression as
20 l2 m 30 a l m = ( 2 2 5 I I m) (2 3 5 a l m )
= (2 5 l m )[(2 I) (3 a )
= 10Im (2I 3a)
(vi) 5 x2 y 15 xy2 :
let 5 x2 y , 14a can be written as
5 x2 y = 5 x x y
15 xy2 = 5 3 x y y
The common factors are given as 5, x, y
Hence, we get Algebraic expression as
5 x2 y 15 xy2 = ( 5 x x y) (5 3 x y y)
= 5 x y(x 3 y )
= 5xy (x 3y)
(vii) 10 a 2 – 15 b2 + 20 c2
let 10 a 2 , 15 b2 , 20 c2 can be written as
10 a 2 = 2 5 a a
15 b2 = 5 3 b b
20 c2 = 2 2 5 c c
The common factors are given as 5
Hence, we get Algebraic expression as
10 a 2 – 15 b2 + 20 c2 = ( 2 5 a a) ( 5 3 b b ) (2 2 5 c c)
= 5 ( 2 a 2 – 3 b2 + 4 c2 )
(viii) – 4 a2 + 4 ab – 4 ca :
let – 4 a2 , 4 ab ,– 4 ca can be written as
– 4 a2 = 2 2 a a
4 ab = 2 2 a b
– 4 ca = 2 2 c a
The common factors are given as 2, 2, a
Hence, we get Algebraic expression as
– 4 a2 4 ab – 4 ca = ( 2 2 a a) ( 2 2 a b ) ( 2 2 c a )
= 2 2 a ( a + b c)
(ix) x 2 y z + x y2 z + x y z2 :
let x 2 y z , x y2 z , x y z2 can be written as
x 2 y z = x x y z
x y2 z = x y y z
x y z2 = x y z z
The common factors are given as z, x, y
Hence, we get Algebraic expression as
x 2 y z + x y2 z + x y z2 = ( x x y z) +( x y y z) + ( x y z z )
= x y z(x + y + z )
= xyz (x + y + z)
(x) a x2 y + b x y2 + c x y z:
let a x2 y , b x y2 , c x y z can be written as
a x2 y = a x x y
b x y2 = b x y y
c x y z = c x y z
The common factors are given as z, x, y
Hence, we get Algebraic expression as
a x2 y + b x y2 + c x y z = ( a x x y) +( b x y y) + ( c x y z )
= x y (ax + by +cz )
= xy (ax + by +cz)


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