## Factorisation

### NCERT Exercise 14.1 Solutions

Q1: Find the common factors of the terms

(i)   12x, 36

(ii)  2y, 22xy

(iii)  14pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6abc, 24ab2, 12a2b

(vi) 16x3, −4x2, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x2y3, 10x3y2, 6x2y2z

(i) 12x = 2 ☓ 2 ☓ 3 ☓ x

36 = 2 ☓ 2 ☓ 3 ☓ 3

The common factors are 2, 2, 3.

And, 2 ☓ 2 ☓ 3 = 12

(ii) 2y = 2 ☓ y

22xy = 2 ☓ 11 ☓ x ☓ y

The common factors are 2, y.

And, 2 ☓ y = 2y

(iii) 14pq = 2 ☓ 7 ☓ p ☓ q

28p2q2 = 2 ☓ 2 ☓ 7 ☓ p ☓ p ☓ q ☓ q

The common factors are 2, 7, p, q.

And, 2 ☓ 7 ☓ p ☓ q = 14pq

(iv) 2x = 2 ☓ x

3x2 = 3 ☓ x ☓ x

4 = 2 ☓ 2

The common factor is 1.

(v) 6abc = 2 ☓ 3 ☓ a ☓ b ☓ c

24ab2 = 2 ☓ 2 ☓ 2 ☓ 3 ☓ a ☓ b ☓ b

12a2b = 2 ☓ 2 ☓ 3 ☓ a ☓ a ☓ b

The common factors are 2, 3, a, b.

And, 2 ☓ 3 ☓ a ☓ b = 6ab

(vi) 16x3 = 2 ☓ 2 ☓ 2 ☓ 2 ☓ x ☓ x ☓ x

−4x2 = −1 ☓ 2 ☓ 2 ☓ x ☓ x

32x = 2 ☓ 2 ☓ 2 ☓ 2 ☓ 2 ☓ x

The common factors are 2, 2, x.

And, 2 ☓ 2 ☓ x = 4x

(vii) 10pq = 2 ☓ 5 ☓ p ☓ q

20qr = 2 ☓ 2 ☓ 5 ☓ q ☓ r

30rp = 2 ☓ 3 ☓ 5 ☓ r ☓ p

The common factors are 2, 5.

And, 2 ☓ 5 = 10

(viii) 3x2y3 = 3 ☓ x ☓ x ☓ y ☓ y ☓ y

10x3y2 = 2 ☓ 5 ☓ x ☓ x ☓ x ☓ y ☓ y

6x2y2z = 2 ☓ 3 ☓ x ☓ x ☓ y ☓ y ☓ z

The common factors are x, x, y, y.

And,

x ☓ x ☓ y ☓ y = x2y2

Q2: Factorise the following expressions

(i) 7x − 42

(ii) 6p − 12q

(iii) 7a2 + 14a

(iv) −16z + 20z3

(v) 20l2m + 30 alm

(vi) 5x2y − 15xy2

(vii) 10a2 − 15b2 + 20c2

(viii) −4a2 + 4ab − 4 ca

(ix) x2yz + xy2z + xyz2

(x) ax2y + bxy2 + cxyz

(i) 7x = 7 ☓ x

42 = 2 ☓ 3 ☓ 7

The common factor is 7.

∴ 7x − 42 = (7 ☓ x) − (2 ☓ 3 ☓ 7) = 7 (x − 6)

(ii) 6p = 2 ☓ 3 ☓ p

12q = 2 ☓ 2 ☓ 3 ☓ q

The common factors are 2 and 3.

∴ 6p − 12q = (2 ☓ 3 ☓ p) − (2 ☓ 2 ☓ 3 ☓ q)

= 2 ☓ 3 [p − (2 ☓ q)]

= 6 (p − 2q)

(iii) 7a2 = 7 ☓ a ☓ a

14a = 2 ☓ 7 ☓ a

The common factors are 7 and a.

∴ 7a2 + 14a = (7 ☓ a ☓ a) + (2 ☓ 7 ☓ a)

= 7 ☓ a [a + 2] = 7a (a + 2)

(iv) 16z = 2 ☓ 2 ☓ 2 ☓ 2 ☓ z

20z3 = 2 ☓ 2 ☓ 5 ☓ z ☓ z ☓ z

The common factors are 2, 2, and z.

∴ −16z + 20z3 = − (2 ☓ 2 ☓ 2 ☓ 2 ☓ z) + (2 ☓ 2 ☓ 5 ☓ z ☓ z ☓ z)

= (2 ☓ 2 ☓ z) [− (2 ☓ 2) + (5 ☓ z ☓ z)]

= 4z (− 4 + 5z2)

(v) 20l2m = 2 ☓ 2 ☓ 5 ☓ l ☓ l ☓ m

30alm = 2 ☓ 3 ☓ 5 ☓ a ☓ l ☓ m

The common factors are 2, 5, l, and m.

∴ 20l2m + 30alm = (2 ☓ 2 ☓ 5 ☓ l ☓ l ☓ m) + (2 ☓ 3 ☓ 5 ☓ a ☓ l ☓ m)

= (2 ☓ 5 ☓ l ☓ m) [(2 ☓ l) + (3 ☓ a)]

= 10lm (2l + 3a)

(vi) 5x2y = 5 ☓ x ☓ x ☓ y

15xy2 = 3 ☓ 5 ☓ x ☓ y ☓ y

The common factors are 5, x, and y.

∴ 5x2y − 15xy2 = (5 ☓ x ☓ x ☓ y) − (3 ☓ 5 ☓ x ☓ y ☓ y)

= 5 ☓ x ☓ y [x − (3 ☓ y)]

= 5xy (x − 3y)

(vii) 10a2 = 2 ☓ 5 ☓ a ☓ a

15b2 = 3 ☓ 5 ☓ b ☓ b

20c2 = 2 ☓ 2 ☓ 5 ☓ c ☓ c

The common factor is 5.

10a2 − 15b2 + 20c2 = (2 ☓ 5 ☓ a ☓ a) − (3 ☓ 5 ☓ b ☓ b) + (2 ☓ 2 ☓ 5 ☓ c ☓ c)

= 5 [(2 ☓ a ☓ a) − (3 ☓ b ☓ b) + (2 ☓ 2 ☓ c ☓ c)]

= 5 (2a2 − 3b2 + 4c2)

(viii) 4a2 = 2 ☓ 2 ☓ a ☓ a

4ab = 2 ☓ 2 ☓ a ☓ b

4ca = 2 ☓ 2 ☓ c ☓ a

The common factors are 2, 2, and a.

∴ −4a2 + 4ab − 4ca = − (2 ☓ 2 ☓ a ☓ a) + (2 ☓ 2 ☓ a ☓ b) − (2 ☓ 2 ☓ c ☓ a)

= 2 ☓ 2 ☓ a [− (a) + b − c]

= 4a (−a + b − c)

(ix) x2yz = x ☓ x ☓ y ☓ z

xy2z = x ☓ y ☓ y ☓ z

xyz2 = x ☓ y ☓ z ☓ z

The common factors are x, y, and z.

∴ x2yz + xy2z + xyz2 = (x ☓ x ☓ y ☓ z) + (x ☓ y ☓ y ☓ z) + (x ☓ y ☓ z ☓ z)

= x ☓ y ☓ z [x + y + z]

= xyz (x + y + z)

(x) ax2y = a ☓ x ☓ x ☓ y

bxy2 = b ☓ x ☓ y ☓ y

cxyz = c ☓ x ☓ y ☓ z

The common factors are x and y.

ax2y + bxy2 + cxyz = (a ☓ x ☓ x ☓ y) + (b ☓ x ☓ y ☓ y) + (c ☓ x ☓ y ☓ z)

= (x ☓ y) [(a ☓ x) + (b ☓ y) + (c ☓ z)]

= xy (ax + by + cz)

Q-3 Factorise

(i) x2 + xy + 8x + 8y

(ii) 15xy − 6x + 5y − 2

(iii) ax + bx − ay − by

(iv) 15pq + 15 + 9q + 25p

(v) z − 7 + 7xy − xyz

(i) x2 + xy + 8x + 8y = x ☓ x + x ☓ y + 8 ☓ x + 8 ☓ y

= x (x + y) + 8 (x + y)

= (x + y) (x + 8)

(ii) 15xy − 6x + 5y − 2 = 3 ☓ 5 ☓ x ☓ y − 3 ☓ 2 ☓ x + 5 ☓ y − 2

= 3x (5y − 2) + 1 (5y − 2)

= (5y − 2) (3x + 1)

(iii) ax + bx − ay − by = a ☓ x + b ☓ x − a ☓ y − b ☓ y

= x (a + b) − y (a + b)

= (a + b) (x − y)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15

= 3 ☓ 5 ☓ p ☓ q + 3 ☓ 3 ☓ q + 5 ☓ 5 ☓ p + 3 ☓ 5

= 3q (5p + 3) + 5 (5p + 3)

= (5p + 3) (3q + 5)

(v) z − 7 + 7xy − xyz = z − x ☓ y ☓ z − 7 + 7 ☓ x ☓ y

= z (1 − xy) − 7 (1 − xy)

= (1 − xy) (z − 7)  