In this page we have *NCERT solutions for class 8 maths chapter 14 Exercise 14.1* for Factorisation.This exercise has questions on finding common factors, Factorisation by regrouping terms. Hope you like them and do not forget to like , social share
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Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14 pq, 28 p

(iv) 2x, 3x², 4

(v) 6 abc, 24ab², 12 a²b

(vi) 16 x³, – 4x², 32x

(vii) 10 pq, 20qr, 30rp

(viii) 3x² y³, 10x³ y²,6 x² y²z

(i) We need to convert each of these terms in simple factors forms and then choose the common factors

12x =2×2×3×x

36=2×2×3×3

So common Factor are 2,2,3

And 2×2×3=12

(ii) 2y=2×y

22xy=2×11×y×x

So common factors are 2,y

And 2×y=2y

(iii) 14 pq= 2×7×p×q

28 p

So common factors are 2,7,p,q

=14pq

(iv) 2x=2×x

3x

4=2×2

There is no common factor other than unity, so common factor= 1

(v) 6abc=2×3×a×b×c

24ab

12a

Common factors are 2, 3, a, b

=6ab

(vi) 16x

-4x

32x=-2×2×2×2×2×x

Common factors are 2, 2, x

=4x

(vii) 10pq=-2×5×p×q

20qr=-2×2×5×q×r

30rp=-2×3×5×p×r

Common factors are 2,5

=10

(viii) 3x

10x

6x

Common factors are x, x, y, y

=x

Factorize the following expressions.

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a² + 14a

(iv) – 16 z + 20 z³

(v) 20 l² m + 30 a l m

(vi) 5 x² y – 15 xy²

(vii) 10 a² – 15 b² + 20 c²

(viii) – 4 a² + 4 ab – 4 ca

(ix) x² y z + x y²z + x y z²

(x) a x² y + b x y² + c x y z²

(i) 7x-42

= (7×x) – (7×6)

=7(x-6)

(ii) 6p-12q

= (6×p) – (6×2q)

=6(p-2q)

(iii) 7a

=(7×a×a) + (2×7×a)

=7a(a+2)

(iv) -16z+ 20z

=4z(5z

(v) 20 l² m + 30 a l m

=(2×2×5×l×m×l) + (2×3×5×a×l×m)

=2lm(10l+15a)

(vi) 5 x²y -15 xy

=(5×x×x×y) + (5×3×x×y×y)

=5x(xy-3y

(vii) 10a² -15 b

=(2×5×a×a) - (5×3×b×b) + (5×2×2×c×c)

=5(2a

(viii) -4a² +4a b-4ca

=(-4×a×a) +(4×b×a) - (4×c×a)

=-4a(a-b+c)

(ix) x²yz + xy

=(x×x×y×z) +(x×y×y×z) +(x×y×z×z)

=xyz(x+y+z)

(x) ax²yz + bxy

=(x×x×y×z×a) +(x×y×y×z×b) +(x×y×z×z×c)

=xyz(ax+by+cz)

Factorize

(i) x² + x y + 8x + 8y

(ii) 15 xy – 6x + 5y – 2

(iii) ax + bx – ay – by

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

(v) z – 7 + 7 x y – x y z

(i) x² + x y + 8x + 8y

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

=(x+y)(x+8)

(ii) 15 xy – 6x + 5y – 2

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

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

(iii) ax + bx – ay – by

= x(a+b) –y(a+b)

=(a+b)(x-y)

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

=15pq+9q +25p+15

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

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

(v) z – 7 + 7 x y – x y z

=z-xyz-7+7xy

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

=(z-7)(1-xy)

- NCERT solutions for class 8 maths chapter 14 Exercise 14.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

Download NCERT solutions Factorisation Exercise 14.1 as pdf - This chapter 14 has total 4 Exercise 14.1,14.2,14.3 and 14.4. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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