Primary School Students‎ > ‎Algebra‎ > ‎

## Factorizing

A polynomial expressed as a product of two or more expressions has each product as its factor.
As x2 + 3x + 2 =(x+1)(x+2).
The expression (x+1) and (x+2) are factors of x2 + 3x + 2 and the process of obtaining (x+1) and (x+2) is called factorization.

## Methods of factorization

There are two methods of factorization
1. Writing out the common factors
2. Grouping and writing out the common factors

### Writing out the common factors

2ax2 + 4a
= 2a(x2 + 2)
Method: Find the term which can be common to all.Divide each term by the common factor and write in brackets. Write the common factor outside the brackets.

### Grouping and writing out the common factors

Factor 2a + 8bx + 8ax2 + 16bx2 + 4b + 4ax
2a + 8bx + 8ax2 + 16bx2 + 4b + 4ax
= 2a + 4ax + 8ax2 + 4b + 8bx + 16bx2
= 2a(1 + 2x + 4x2) + 2b(1 + 2x + 4x2)
Here (1 + 2x + 4x2) is also a common factor so follow the first method of writing out the common factors.
= (1 + 2x + 4x2) (2a + 2b)

Identify which term is common to some. Group those terms. Follow the method of writing out the common factors.

## Factorizing ax2 + bx + c, A quadratic trinomial

Steps:Factorize: 2x2 − 2x − 4
1. Find the factors of a×c
a×c = 2×−4 = −8.
Factors are − 2,2,2 or − 4,2 or 4,− 2 or −8,1 or 8,−1.

2. Find the sum or difference of any two factors which is equal to b.
b = −2 = −4 + 2

3. Substitute the sum or difference for b.
2x2 − 2x − 4
= 2x2 +(−4+2)x − 4
= 2x2 −4x + 2x − 4

4. Solve by grouping and writing out the common factors.
2x2 −4x + 2x − 4
= 2x(x − 2) + 2(x − 2)
= (x − 2) (2x + 2)
= 2(x − 2)(x + 1)

## Factorizing a2 − b2

a2 − b2 = (a + b)(a − b)

## Factorizing a3 ± b3

a3 + b3 = (a + b)(a2 − ab + b2)
a3 − b3 = (a − b)(a2 + ab + b2)

## Factorization

The process of writing an expression as the product of two or more expressions is called factorization. Each expression occurring in the product is called factor of the given expression.

## Process of factorization

1. Look for the common factor in the expression.
2. The factor can be monomial or an expression.
3. If the factor look like a monomial. Take out the HCF of the terms. Treat it as one of the factor.
4. If the factor looks like an expression. Takeout the expression as the common term.

### Find the factors of 15ab + 3ac.

Here a looks to be the monomial common in both the terms.
The HCF of 15ab and 3ac is 3a.
15ab + 3ac = 3a×5b + 3a×c
3a(5b + c)
Hence the factors are 3a and 5b + c

### Find the factors of 3a(5b + c) + 4d(5b + c)

Here (5b + c) looks to be the factor.
(5b + c)(3a + 4d)
The factors are (5b + c) and (3a + 4d).

To factor (a2 - b2) we use
(a2 - b2) = (a + b)(a - b)

### To factor (ax2 + bx + c)

(ax2 + bx + c) = a(x + A)(x + B)
(ax2 + a(A + B)x + aAB)
We have to find b such that b = a(A + B) and c = aAB
i.e. b/a = (A + B) and c/a = AB
We have to find two numbers such that their sum is b/a and product is c/a.
When a = 1 then the expression is of the form (x2 + bx + c).