Mathematics – Class 8 – Chapter 14 – Factorisation – Exercise 14.1 – NCERT Exercise Solution

1. Find the common factors of the given terms.

(i) 12x, 36

Solution:

Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors are 2, 2, 3

Hence, common factors are 2×2×3 = 12

(ii) 2y, 22xy

Solution:

Factors of 2y and 22xy

2y = 2 × y

22xy = 2×11× x ×y

Common factors = 2, y

Hence common factors are 2 × y = 2y

(iii) 14 pq, 28p²q²

Solution:

Factors of 14pq and 28p²q²

14pq = 2 x 7 x p x q

28p²q² = 2 x 2 x7 x p x p x q x q

Common factors are 2, 7 , p , q

Hence, common factors are 2 x 7 x p x q = 14pq

(iv) 2x, 3x², 4

Solution:

Factors of 2x, 3x²and 4

2 x = 2 × x

3 x 2= 3 × x × x

4 = 2 × 2

Hence, common factor is 1.

(v) 6 abc, 24ab², 12a²b

Solution:

Factors of 6abc, 24ab² and 12a²b

6abc = 2 × 3 × a × b × c

24ab² = 2 × 2 × 2 × 3 × a × b × b

12 a² b = 2 × 2 × 3 × a × a × b

Common factors are 2, 3, a, b

Hence, common factors are 2 × 3 × a × b = 6ab

(vi) 16 x³, – 4x², 32 x

Solution:

Factors of 16x³ , -4x²and 32x

16 x³ = 2 × 2 × 2 × 2 × x × x × x

– 4x² = -1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

Common factors are 2,2, x

Hence, common factors are 2×2×x = 4x

(vii) 10 pq, 20qr, 30 rp

Solution:

Factors of 10 pq, 20qr and 30rp

10 p q = 2 × 5 × p × q

20 q r = 2 × 2 × 5 × q × r

30 r p= 2× 3 × 5 × r × p

Common factors are 2, 5

Hence, common factors are 2×5 = 10

(viii) 3x²y³, 10x³y², 6x²y²z

Solution:

Factors of 3x²y³, 10x³y² and 6x²y²z

3 x² y³ = 3 × x × x × y × y × y

10x³ y² = 2 × 5 × x × x × x × y × y

6x²y²z = 3 × 2 × x × x × y × y × z

Common factors are x², y²

Hence, common factors are x² × y² = x²y²

2.Factorise the following expressions

(i) 7x–42

Solution:

Take 7 as common

= 7 (x – 6)

(ii) 6p–12q

Solution:

Take 6 as common

= 6(p – 2q)

(iii) 7a²+ 14a

Solution:

Take 7a as common

=7a (a + 2)

(iv) -16z+20 z³

Solution:

Take 4z as common

= 4z ( -4 + 5 z²)

(v) 20l²m+30alm

Solution:

Take 10 l m as common

= 10 l m (2 l + 3a)

(vi) 5x²y-15xy²

Solution:

Take 5xy as common

= 5xy (x – 3y)

(vii) 10a²-15b²+20c²

Solution:

Take 5 as common

= 5 (2 a² – 3 b² + 4 c²)

(viii) -4a²+4ab–4 ca

Solution:

Take 4a as common

= 4a (-a + b – c)

(ix) x²yz+xy²z +xyz²

Solution:

Take xyz as common

= xyz ( x + y + z)

(x) ax²y+bxy²+cxyz

Solution:

Take xy as common

= xy (ax + by + z)

3. Factorise.

(i) x²+xy+8x+8y

Solution:

= (x²+xy) + (8x+8y)

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

= (x + y)  (x + 8)

Hence, required factors: (x + y) (x + 8)

(ii) 15xy–6x+5y–2

Solution:

= (15xy–6x) + (5y–2)

= 3x (5y – 2) + (5y – 2)

= (5y – 2) (3x +1)

Hence, required factors: (5y – 2) (3x +1)

(iii) ax+bx–ay–by

Solution:

= (ax – ay) – (bx –by)

= a (x – y) – b (x – y)

= (a – b) (x – y)

Hence required factors: (a – b) (x – y)

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

Solution:

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

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

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

Hence, required factors: (5p + 3) (3q + 5)

(v) z–7+7xy–xyz

Solution:

= (-xyz + 7xy) + (z – 7)

= -xy (z – 7) + (z – 7)

= (-xy + 1) (z – 7)

Hence, required factors: (-xy + 1) (z – 7)