# Factorisation Chapter 14 Class 8 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 8 Maths Chapter 14′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 8 Maths Chapter 14 Exercise Solution’ using the download button.

### Chapter 14:Factorisation

#### 14.1.1 Factors of natural numbers

You will remember what you learned about factors in Class VI. Let us take a natural number, say 30, and write it as a product of other natural numbers, say
30 = 2 × 15
= 3 × 10 = 5 × 6

Thus, 1, 2, 3, 5, 6, 10, 15, and 30 are the factors of 30. Of these, 2, 3, and 5 are the prime factors of 30 (Why?)

A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form.

The prime factor form of 70 is 2 × 5 × 7.
The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.

Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter.

#### 14.1.2 Factors of algebraic expressions

We have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term, 5xy has been formed by the factors 5, x, and y, i.e., 5xy = 5× x × y

Observe that the factors 5, x, and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x, and y are ‘prime’ factors of 5xy. In algebraic expressions,

we use the word ‘irreducible’ in place of ‘prime’. We say that 5 × x × y is the irreducible form of 5xy. Note 5 × (XY) is not an irreducible form of 5xy, since the factor XY can be further expressed as a product of x and y, i.e., XY = x × y

### NCERT Solutions Class 8 Maths Chapter 14 Factorisation

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 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

3x2= 3×x×x

4 = 2×2

Common factors of 2x, 3xand 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 ab = 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 16x, -4x2and 32x

16 x= 2×2×2×2×x×x×x

– 4x2 = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x, – 4xand 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

10xy2 = 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