# Factorisation

Notes of chapter: Factorisation are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 8. (i) The prime factors of a natural number are prime numbers when multiply together we will get that number.

Eg.- Find the prime factors of 40.

Ans-

40 = 2 × 2 × 2 × 5

Hence, 2 and 5 are prime factors of 40.

(ii) The factors of an algebraic expression are terms when multiply together we will get that algebraic expression.

Eg:-Find the factors of 2x y.

Ans-

It is an algebraic expression.

2xy is a term.

Factors of term 2xy =2,x,y

[Note – 1 is factor of every term. We do not write 1 as a separate factor unless it is specially required.]

Types of factors of algebraic expressions-

(a) Irreducible factors of algebraic expressions are the factors which can not be factorised further.

Eg:-

(A) Find the irreducible factors of 2xy.

Irreducible factors of 2xy =2,x,y

2, x and y all are irreducible factors because we can not reduce them.

(B) Find the irreducible factors of 5x(y + 2).

Irreducible factors of 5x(y + 2) = 5 × x × (y + 2)

5,x and (y + 2 ) are irreducible factors.

(b) Reducible factors of algebraic expressions are the factors which can be factorised further.

Eg:- 2xy is a reducible factors.

(2)Factorisation is a method to do the factors as a product of the number or algebraic expression.

Eg:-

(i) Factorise a number 35.

Ans-

35 = 5 × 7

Hence, 1, 5, 7and 35 are factors of 35.

(ii)Factorise an algebraic expression 4 x2.

4 x2 = 2 × 2 × x × x

Hence, 1, 2, x are factors of 4x2.

Methods of factorisation

(i) Method of common factors

(a)Factorise 3x + 9

Step 1-

Find irreducible factors of each term.

3x = 3 × x

9 = 3 × 3

Therefore,

3x + 9 = (3× x) + (3 × 3)

Step 2-

Find the common factor of terms.

In above factors only 3 is a common factor.

Step 3-

By using distributive law, take out the common factors of factors.

(3× x) + (3 × 3) = 3(x + 3)

Step 4-

Hence,

3x + 9 = 3(x + 3)

3 and (x + 3) are the factors of the expression 3x + 9.

(b) Factorise 3xy + 9x

Step 1 –

Find irreducible factors of each term.

3xy = 3 × x × y

9x = 3×  3 × x

Therefore,

3xy + 9x = (3 × x × y) + (3 × 3 × x)

Step 2-

Find the common factor of terms.

In above factors only 3 and x are common factors.

Step 3-

By using distributive law, take out the common factors of terms.

(3× x × y) + (3× 3× x) = 3x (y + 3)

Step 4-

Hence,

3xy + 9x = 3x(y + 3)

3x and (y + 3) are the factors of the expression 3x y+ 9x.

(ii) Method of regrouping terms

Regrouping means rearranging the terms in different groups in such a way that each group must has one common factor.

Eg:- Find the factor of expression 4x y + 5 + 2x + 10y

Step 1-

Regroup the expressions in such a way that each group must have one common factor.

4x y + 5 + 2x + 10y

Regroup- (4xy + 2x )+ (5 + 10y)

Step 2 –

Find irreducible factors of each term.

4xy = 2 × 2 × x × y

2x = 2×x

10y = 5 × 2 ×  y

5 = 5

Therefore

4xy + 2x + 5 + 10y = (4  x  y + 2  x) + (5 + 5 × 2 × y) (iii) Factorisation using identities

(a) (a + b)2 = a 2+ 2ab + b2

(b) (a – b)2 = a 2– 2ab + b2

(c) (a + b) (a – b) = a2– b2

(d) (x + a ) (x + b) = x2 + (a + b)x + ab

Steps to use identities-

(a) (a + b)2 = a 2+ 2ab + b2

Eg:- Factorise x2 + 6x + 9

Ans-

x2 + 6x + 9

In above expression

a = x, b = 3, 2ab = 2x × 3

Put values in the identity (a + b)2 = a 2+ 2ab + b2

Therefore,

x2 + 6x + 9 = (x + 3)2 Required factorisation

(b) (a – b)2 = a 2– 2ab + b2

Eg:- Factorise x2 – 6x + 9

Ans-

x2 – 6x + 9

In above expression

a = x, b = 3, 2ab = 2x × 3

Put values in the identity (a – b)2 = a 2– 2ab + b2

Therefore,

x2 – 6x + 9 = (x – 3)2 Required factorisation

(c) (a + b) (a – b) = a2– b2

Eg:- Factorise x2 – 16

Ans-

In the above expression

a2 = x2, b2 = 42

Put values in the identity (a + b) (a – b) = a2– b2

x2 – 22 = (x + 2) (x – 2) Required factorisation      Helping Topics

NCERT solutions, class 8

Practice sheet, class 8