# Factoring Algebraic Expressions

Factoring (factorising or factorizing)  is the process of splitting an algebraic expression and writing it as a product of its factors. Factors are building blocks of an expression, like how numbers can be broken down into prime factors.

We factor expressions to get a simplified version, which is easier to work with while finding values of an unknown variable.

As we know, 16 can be factored as 1 x 16, 2 x 8, and 4 x 4. Thus, 1, 2, 4, 8, 16 are the factors of 16.

Similarly, algebraic expressions can be factored too.

The expression, ${x^{2}+2x}$ can be factored as x(x + 2).Thus, x and x + 2 are the factors of ${x^{2}+2x}$.

It is thus the reverse of expanding brackets using the distributive property.

There are many ways to factor algebraic expressions based on their types:

## Methods

### By Factoring Common Terms

Let us factor the expression (${-5x^{2}+20x}$).

First, we factor each term of ${-5x^{2}+20x}$,

${-1\times 5\times x\times x+5\times 2\times 2\times x}$

Now, taking out the highest common factor (here, 5x), we get

${5\times x\left( -1\times x+2\times 2\right)}$

= 5x(-x + 4)

Thus, the expression is factored into single brackets using the common factor method.

### By Factoring Middle-Terms

This method usually factors quadratic expressions of the form ${ax^{2}+bx+c}$.

Let us factor the expression ${x^{2}+7x+10}$

Here, 7x (middle term) has been split into two numbers such that their product will be the product of the first and third terms in the given expression.

Thus, the second term is 7x = (2 + 5)x, and 2 x 5 = 10 is the third term.

Now, the given expression ${x^{2}+7x+10}$ can be written as ${x^{2}+2x+5x+10}$

By grouping the terms and taking out the common factors, we get

= x(x + 2) + 5(x + 2)

Again, by taking out the common term (x + 2) and keeping the remaining terms in the brackets, we get

= (x + 2)(x + 5)

### By Regrouping

However, some expressions require regrouping before they can be factored into  double brackets:

As in the case with the expression, 12x + y – xy -12

Here, we observe that 12 is the common factor of the first and last term, whereas y is the common factor of the second and third term.

Thus, the terms can be regrouped as 12x – 12 +y – xy

= 12x – 12 – xy + y

Now, taking out the common factors and keeping the remaining ones in the brackets, we get

12(x – 1) + y(x – 1)

= (x – 1)(12 + y)

Thus, we factor the expressions into double brackets by splitting the middle terms into two parts and regrouping them.

### By Using Identities

Another method of factoring algebraic expressions is done by using the following identities.

Let us factor the expression ${9x^{2}-16y^{2}}$ using identities.

As we can see, there are no common factors for the terms. However, ${9x^{2}}$ and ${16y^{2}}$ are the perfect squares.

Hence, the given expression is expressed as the difference of two squares.

Using the identity of ${a^{2}-b^{2}}$, we get

(3x + 4y)(3x – 4y)

Thus, by using the identities, we can factor the expressions.

## Solved Examples

Factorise: ${x^{2}-10x+25}$

Solution:

In the given expression, there are no common factors for all three terms.
However, an identity ${\left( a-b\right) ^{2}}$ = ${a^{2}-2ab+b^{2}}$,matches the given algebraic expression.
${x^{2}-10x+25}$ can be written as ${x^{2}-2\times 5\times x+5^{2}}$
= ${\left( x-5\right) ^{2}}$
Thus, the factors are ${\left( x-5\right) ^{2}}$ or (x – 5)(x – 5).

Find the factors of the following expressions:
a) ${15x^{2}y^{3}-9x^{4}y^{2}}$
b) ${3y^{2}+20y+12}$

Solution:

a) The given expression is ${15x^{2}y^{3}-9x^{4}y^{2}}$
= ${3\times 5\times x\times x\times y\times y\times y}$ – ${3\times 3\times x\times x\times x\times x\times y\times y}$Â
= ${3\times x\times x\times y\times y}$ ${\left( 5y-3x^{2}\right)}$
= ${3x^{2}y^{2}\left( 5y-3x^{2}\right)}$
b) The given expression is ${3y^{2}+13y+12}$
Here, 13y has been split into two numbers so that their product will be the product of ${3y^{2}}$ and 12.
As we know, 3 x 12 = 36 = 4 x 9
Thus, 13y = (4 + 9)y
The given expression can be written as ${3y^{2}+4y+9y+12}$
Now, by grouping them, we get
(${3y^{2}}$ + 4y) + (9y + 12)
= y(3y + 4) + 3(3y + 4)
= (3y + 4)(y + 3)