Class 8 | NCERT Solution Maths Chapter 14 | Factorisation | Exercise 14.1

Exercise 14.1 Page No: 208

1. Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28p2q2

(iv) 2x, 3x2, 4

(v) 6 abc, 24ab2, 12a2b

(vi) 16 x3, – 4x2 , 32 x

(vii) 10 pq, 20qr, 30 rp

(viii) 3x2y3 , 10x3y2 , 6x2y2z

Solution:

(i) Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

Common factors of 12x and 36 are 2, 2, 3

and , 2×2×3 = 12

(ii) Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

Common factors of 2y and 22xy are 2, y

and ,2×y = 2y

(iii) Factors of 14pq and 28p2q

14pq = 2x7xpxq

28p2q = 2x2x7xpxpxq

Common factors of 14 pq and 28 p2q are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x2and 4

2x = 2×x

3x2= 3×x×x

4 = 2×2

Common factors of 2x, 3x2 and 4 is 1.

(v) 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 of 6 abc, 24ab2 and 12a2b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) 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 of 16 x3 , – 4x2 and 32x are 2,2, x

and, 2×2×x = 4x

(vii) Factors of 10 pq, 20qr and 30rp

10 pq = 2×5×p×q

20qr = 2×2×5×q×r

30rp= 2×3×5×r×p

Common factors of 10 pq, 20qr and 30rp are 2, 5

and, 2×5 = 10

(viii) Factors of 3x2y3 , 10x3y2 and 6x2y2z

3x2y3 = 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 of 3x2y3, 10x3y2 and 6x2y2z are x2, y2

and, x2×y2 = x2y2

2.Factorise the following expressions

(i) 7x–42

(ii) 6p–12q

(iii) 7a2+ 14a

(iv) -16z+20 z3

(v) 20l2m+30alm

(vi) 5x2y-15xy2

(vii) 10a2-15b2+20c2

(viii) -4a2+4ab–4 ca

(ix) x2yz+xy2z +xyz2

(x) ax2y+bxy2+cxyz

Solution:

(vii) 10a2-15b2+20c2

10a2 = 2×5×a×a

– 15b2 = -1×3×5×b×b

20c2 = 2×2×5×c×c

Common factor of 10 a2 , 15b2 and 20c2 is 5

10a2-15b2+20c2 = 5(2a2-3b2+4c2 )

(viii) – 4a2+4ab-4ca

– 4a2 = -1×2×2×a×a

4ab = 2×2×a×b

– 4ca = -1×2×2×c×a

Common factor of – 4a2 , 4ab , – 4ca are 2, 2, a i.e. 4a

So,

– 4a2+4 ab-4 ca = 4a(-a+b-c)

(ix) x2yz+xy2z+xyz2

x2yz = x×x×y×z

xy2z = x×y×y×z

xyz2 = x×y×z×z

Common factor of x2yz , xy2z and xyz2 are x, y, z i.e. xyz

Now, x2yz+xy2z+xyz2 = xyz(x+y+z)

(x) ax2y+bxy2+cxyz

ax2y = a×x×x×y

bxy2 = b×x×y×y

cxyz = c×x×y×z

Common factors of a x2y ,bxy2 and cxyz are xy

Now, ax2y+bxy2+cxyz = xy(ax+by+cz)

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

Solution: