Courses
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
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
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)