In this page we have *Ncert Solutions for Class 8 Mathematics Chapter 14 - Factorization Chapter 14 * for
Exercise 14.1 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

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

1) 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

2) 2y=2×y

22xy=2×11×y×x

So common factors are 2,y

And 2×y=2y

3) 14 pq= 2×7×p×q

28 p

So common factors are 2,7,p,q

=14pq

4) 2x=2×x

3x

4=2×2

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

5) 6abc=2×3×a×b×c

24ab

12a

Common factors are 2, 3, a, b

=6ab

6) 16x

-4x

32x=-2×2×2×2×2×x

Common factors are 2, 2, x

=4x

7) 10pq=-2×5×p×q

20qr=-2×2×5×q×r

30rp=-2×3×5×p×r

Common factors are 2,5

=10

8) 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

1) 7x-42

= (7×x) – (7×6)

=7(x-6)

2) 6p-12q

= (6×p) – (6×2q)

=6(p-2q)

3) 7a

=(7×a×a) + (2×7×a)

=7a(a+2)

4) -16z+ 20z

=4z(5z

5) 20 l² m + 30 a l m

=(2×2×5×l×m×l) + (2×3×5×a×l×m)

=2lm(10l+15a)

6) 5 x²y -15 xy

=(5×x×x×y) + (5×3×x×y×y)

=5x(xy-3y

7) 10a² -15 b

=(2×5×a×a) - (5×3×b×b) + (5×2×2×c×c)

=5(2a

8) -4a² +4a b-4ca

=(-4×a×a) +(4×b×a) - (4×c×a)

=-4a(a-b+c)

9) x²yz + xy

=(x×x×y×z) +(x×y×y×z) +(x×y×z×z)

=xyz(x+y+z)

10) 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

1) x² + x y + 8x + 8y

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

=(x+y)(x+8)

2) 15 xy – 6x + 5y – 2

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

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

3) ax + bx – ay – by

= x(a+b) –y(a+b)

=(a+b)(x-y)

4) 15 pq + 15 + 9q + 25p

=15pq+9q +25p+15

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

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

5) z – 7 + 7 x y – x y z

=z-xyz-7+7xy

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

=(z-7)(1-xy)

Download NCERT solutions Factorization Exercise 14.1 as pdf

**Notes**-
**Worksheets** -
**Ncert Solutions**

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

Thanks for visiting our website.

**DISCLOSURE:** THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY **DISCLOSURE** FOR MORE INFO.