# CLASS-8ALGEBRAIC FACTORIZATION

FACTORIZATION

If a given algebraic expression can be expressed as a product of two or more expressions than those expressions are called factors of the given expression.

Example.1) 20x + 16 = 4 (5x + 4), here 4 & (5x + 4) are two factors of 20x + 16

Example.2) x²- y²= (x+y)(x-y), so (x +y) & (x-y) are two factors of x²- y²

Example.3) x²- 2x + 15 = (x + 3)(x – 5), here (x + 3) & (x – 5) are two factors of x²- 2x + 15

Example.4)  x²+ 5x + 6 = (x+2)(x+3), here (x+2) & (x+3) are two factors of  x²+ 5x + 6

Methods Of Factorization

To resolve an expression into its factors is called factorization. The method we use to factorize an expression depends on the type of expression. In this chapter, we should take a look at a few methods of factorization which are used mostly.

Taking out common factors –

We should follow the steps when all the terms of a polynomial have common factors are discussed below

Step.1) Find the highest common factors (HCF) of the terms by (1) finding the HCF of the numerical coefficients of the terms, (2) finding the highest power of each variable common to the terms, and (3) multiplying all the common factors.

Step.2) Divide each term of the polynomial by this common factor.

Step.3) Write the quotient within brackets and the common factor outside the brackets.

There are some example are given below for your better understanding –

Example.1)  Factorize 4ab – 8b²

Ans.)   4ab = 4 X a X b ,   8b² = 4 X 2 X b X b

The HCF of the numerical coefficients of the terms = 4

The highest power of ‘b’ common to all the terms = b

So, the HCF of the terms = 4b

4ab        8b²

Dividing the expression by the common factor, ------ - ------ = a – 2b

4b        4b

So, the desire result is => 4ab – 8b² =  4b(a – 2b)           (Ans.)

Example.2)  Factorize 9a²b⁴ + 27a²bᶟ - 45aᶟb⁴

Ans.)  9a²b² = 3 X 3 X a X a X b X b X b X b,  27a²bᶟ = 3 X 3 X 3 X a X a X b X b X b, 45aᶟb⁴ = 3 X 3 X 5 X a X a X b X b X b X b

The HCF of the numerical coefficients is = 9

The highest common power of ‘a’ is = a X a

The highest common power of ‘b’ is = b X b X b = bᶟ

So, the HCF of the terms =  9a²bᶟ

Now, we have to divide each term by

9a²b⁴       27a²bᶟ       45aᶟb⁴

Now, 9a²b²+ 27a²bᶟ - 45aᶟb⁴ = 9a²bᶟ (-------- + -------- - --------)

9a²bᶟ        9a²bᶟ        9a²bᶟ

=   9a²bᶟ ( b + 3 – 5ab )          (Ans.)

Example.3)  Factorize  3(a+b)²+ 9(a²+ ab)

Ans.)   3(a+b)² = 3 X (a+b) X (a+b)

9(a²b+ ab²) = 3 X 3 X a X b X (a+b)

The HCF of the numerical coefficient is = 3

The highest power of the binomial (a+b) common to both the terms = a + b

The HCF of the terms = 3(a+b)

So,  3(a+b)² + 9(a²b+ ab²)

3(a+b)²         9(a²b+ ab²)

=  3(a+b) { ----------- + ------------- }

3(a+b)             3(a+b)

=   3(a+b) {(a+b) + 3ab}           (Ans.)