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## NCERT Class 8 Maths Textbook Chapter 14 With Answer Book PDF Free Download

### Chapter 14:Factorisation

#### 14.1 Introduction

#### 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

Author | NCERT |

Language | English |

No. of Pages | 14 |

PDF Size | 221 KB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### 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, 28p ^{2}q^{2}**

**(iv) 2x, 3x ^{2}, 4**

**(v) 6 abc, 24ab ^{2}, 12a^{2}b**

**(vi) 16 x ^{3}, – 4x^{2} , 32 x**

**(vii) 10 pq, 20qr, 30 rp**

**(viii) 3x ^{2}y^{3} , 10x^{3}y^{2} , 6x^{2}y^{2}z**

**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 28p^{2}q^{2}

14pq = 2x7xpxq

28p^{2}q^{2} = 2x2x7xpxpxqxq

Common factors of 14 pq and 28 p^{2}q^{2} are 2, 7 , p , q

and, 2x7xpxq = 14pq

(iv) Factors of 2x, 3x^{2}and 4

2x = 2×x

3x^{2}= 3×x×x

4 = 2×2

Common factors of 2x, 3x^{2 }and 4 is 1.

(v) Factors of 6abc, 24ab^{2} and 12a^{2}b

6abc = 2×3×a×b×c

24ab^{2} = 2×2×2×3×a×b×b

12 a^{2 }b = 2×2×3×a×a×b

Common factors of 6 abc, 24ab^{2} and 12a^{2}b are 2, 3, a, b

and, 2×3×a×b = 6ab

(vi) Factors of 16x^{3 }, -4x^{2}and 32x

16 x^{3 }= 2×2×2×2×x×x×x

– 4x^{2} = -1×2×2×x×x

32x = 2×2×2×2×2×x

Common factors of 16 x^{3 }, – 4x^{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 3x^{2}y^{3} , 10x^{3}y^{2} and 6x^{2}y^{2}z

3x^{2}y^{3} = 3×x×x×y×y×y

10x^{3 }y^{2} = 2×5×x×x×x×y×y

6x^{2}y^{2}z = 3×2×x×x×y×y×z

Common factors of 3x^{2}y^{3}, 10x^{3}y^{2} and 6x^{2}y^{2}z are x^{2}, y^{2}

and, x^{2}×y^{2} = x^{2}y^{2}

NCERT Class 8 Maths Textbook Chapter 14 With Answer Book PDF Free Download