class 8 maths chapter 14 exercise 14.1 (1)

1. Find the common factors of the given terms.

(i) 12x, 36 (ii) 2y, 22xy (iii) 14 pq, 28p²q² (iv) 2x, 3x², 4 (v) 6abc, 24ab², 12a²b (vi) 16x³, –4x², 32x (vii) 10 pq, 20qr, 30rp (viii) 3x²y³, 10x³y², 6x²y²z.

Solution to class 8 maths chapter 14 exercise 14.1 (1).

(i) 12x=2×2×3×x

36=2×2×3×3

Hence, the common factors are 2,2 and 3=2×2×3=12

(ii) 2y=2×y

22xy=2×11×x×y

Hence, the common factors are 2 and y =2×y=2y

(iii) 14pq=2×7×p×q

28p²q²=2×2×7×p×p×q×q

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

(iv) 2x=2×x×1

3x²=3×x×x×1

4=2×2×1

Hence 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

Hence, the common factors are 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

Hence, the common factors are 2×2×x=4x

(vii) 10pq=2×5×p×q

20qr =2×2×5×q×r

30rp=2×3×5×r×p

Hence, the common factors are 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

Hence, the common factors are x×x×y×y=x²y²

class 8 maths chapter 14 exercise 14.1 (2)

2. Factorize the following expressions.

(i) 7x-42 (ii) 6p-12q (iii) 7a²+14a (iv) -16z+20z³

(v) 20l²m+30alm (vi) 5x²y-15xy² (vii) 10a²-15b²+20c²

(viii) -4a²+4ab-4ca (ix) x²yz+xy²z+xyz² (x) ax²y+bxy²+cxyz

Solution to class 8 maths chapter 14 exercise 14.1 (2).

(i) 7x-42=7x – 2×3×7

Taking common factors from each term,

=7(x-2×3)

=7(x-6)

(ii) 6p – 12q=2×3×p – 2×2×3×q

Taking common factors from each term,

=2×3(p – 2q)

=6(p – 2q)

(iii) 7a²+14a=7×a×a + 2×7×a

Taking common factors from each term,

=7×a(a+2)

=7a(a+2)

(iv) -16z + 20z³ = (-1)×2×2×2×2×z+2×2×5×z×z×z

Taking common factors from each term,

=2×2×z(- 2×2+5×z×z)

=4z(- 4+5z²)

(v) 20l²m+30alm=2×2×5×l×l×m+2×3×5×a×l×m

Taking common factors from each term,

=2×5×l×m(2×l+3×a)

=10im(2l+3a)

(vi) 5x²y – 15xy² = 5×x×x×y + 3×5×x×y×y

Taking common factors from each term,

=5×x×y(x-3y)

=5xy(x – 3y)

(vii) 10a² – 15b²+20c² = 2×5×a×a – 3×5×b×b + 2×2×5×c×c

Taking common factors from each term,

=5(2×a×a – 3×b×b + 2×2×c×c)

=5(2a² – 3b² +4c²)

(viii) – 4a²+4ab – 4ca = (-1)×2×2×a×a + 2×2×a×b – 2×2×c×a

Taking common factors from each term,

= 2×2×a(- a + b – c)

= 4a(- a+b+c)

(ix) x²yz+xy²z+xyz² = x×x×y×z+x×y×y×z+x×y×z×z

Taking common factors from each term,

=x×y×z(x+y+z)

=xyz(x+y+z)

(x) ax×²y+bxy²+cxyz= a×x×x×y+b×x×y×y+c×x×y×z

Taking common factors from each term,

=x×y(a×x+b×y+c×z)

=xy(ax+by+cz)

class 8 maths chapter 14 exercise 14.1 (3)

Factorize:

(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 to class 8 maths chapter 14 exercise 14.1 (2).

(i) x²+xy+8x+8y= x(x+y)+8(x+y)

=(x+y)(x+8)

(ii) 15xy-6x+5y-2= 3x(5y-2) + 1(5y-2)

=(5y-2)(3x+1)

(iii) ax+bx-ay-by=(ax+bx) – (ay+by)

=x(a+b) – y(a+b)

=(a+b)(x-y)

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

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

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

(v) z-7+7xy-xyz=7xy – 7 – xyz+z

=7(xy-1)-z(xy-1)

=(xy-1)(7-z)=(-1)(1-xy)(-1)(z-7)

=(1-xy)(z-7)

## class 8 maths chapter 14

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class 8 maths chapter 14 deals with Factorisation.

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