NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.1

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.1

NCERT Solutions Class 8th Maths Chapter 14 Factorisation Exercise 14.1

?Chapter – 14?

✍Factorisation✍

?Exercise 14.1?

1. Find the common factors of the given terms.

(i) 12x, 36
(ii) 2y, 22xy
(iii) 14 pq, 28p2q2
(iv) 2x, 3×2, 4
(v) 6 abc, 24ab2, 12a2b
(vi) 16 x3, – 4×2 , 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 28p2q2
14pq = 2x7xpxq
28p2q2 = 2x2x7xpxpxqxq
Common factors of 14 pq and 28 p2q2 are 2, 7 , p , q
and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x2and 4
2x = 2×x
3×2= 3×x×x
4 = 2×2
Common factors of 2x, 3×2 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 16×3 , -4x2and 32x
16 x3 = 2×2×2×2×x×x×x
– 4×2 = -1×2×2×x×x
32x = 2×2×2×2×2×x
Common factors of 16 x3 , – 4×2 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
10×3 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:

(i) 7x = 7x x
42 = 2 x 3 x 7
The common factor is 7
= 7 x 42 (7 x x) – (2 x 3 x 7) = 7(x – 6)

(ii) 6p = 2 x 3 x p
12 q = 2 × 2 × 3 × q
The common factors are 2 and 3
6p-12 q = (2x 3 x p) – (2 x 2 × 3 × q)
= 2 × 3 [p- (2 x q)]
= 6(p = 2q)

(iii) 7 a² = 7×axa
14 a= 2 x 7 x a
The common factors are 7 and a
= 7a² + 14 a (7 xa xa) + (2x 7x a)
= 7 xa [a + 2] = 7a (a + 2)

(iv) 16 = 2 x 2x2x2x=
20 = 2 x 2 x 5x = x= x=
The common factors are 2, 2, and z.
-16z +20z = -(2x2x2x2x) + (2×2 x 5 xzxzxz)
= (2x 2 × =) – (2 x 2) + (5 x = x =)]
= 4=(-4+5=²)

(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: