# Factoring Difference of Squares

A binomial in the form \(a^2 – b^2\) can be factored as \(\left( {a + b} \right)\left( {a – b} \right)\) or \((\left( {a – b} \right)\left( {a + b} \right)\). The order of the factors doesn’t matter because multiplication is commutative. This is known as the difference of two squares because it represents the difference between two squared quantities. Specifically, it involves subtracting the square of the second quantity \({b^2}\) from the square of the first quantity \({a^2}\), resulting in \({a^2} – {b^2}\).

**Example 1**: Factor the difference of two squares.

$${x^2} \,-\, 49$$

Notice that the first term is a square because the exponent of the variable \(x\) is \(2\). Similarly, the second term is a square, as it can be written as the square of an integer, specifically \(49 = 7^2\). Therefore, we can now express the original problem as the difference of two squares.

$${x^2} \,- \,{7^2}$$

Comparing what we have so far with the difference of squares formula,

$${a^2} \,- \,{b^2} = \left( {a + b} \right)\left( {a\, – \,b} \right)$$

It is obvious to see that \(a=x\) and \(b=7\). Therefore, we have

$${x^2}\, -\, {7^2} = \left( {x + 7} \right)\left( {x – 7} \right)$$

**Example 2**: Factor the difference of two squares.

$${y^2} \,- \,121$$

The first term is a square because of the exponent \(2\). The second term is also a square because it can be expressed as the square of a positive integer: \(121 = 11^2\). Since the conditions are satisfied, we can proceed with factoring. Here, \(a = y\) and \(b = 11\).

$$\eqalign{

{y^2} – 121 &= {y^2} – {11^2} \cr

&= \left( {y + 11} \right)\left( {y – 11} \right) \cr} $$

Just to verify, let’s check if our factors are correct. We will use the FOIL method to multiply the factors and see if we get back to the original problem.

$$\eqalign{

\left( {y + 11} \right)\left( {y – 11} \right) &= {y^2} + 11y – 11y – 121 \cr

&= {y^2} – 121 \cr} $$

Yes, we did! That means we have correctly factored out the binomial.

**Example 3**: Factor the binomial below.

$$4{x^2} \,- \,9{y^2}$$

At first glance, this might not seem like a difference of two squares. However, observe that the first term can be written as \(4{x^2} = {\left( {2x} \right)^2}\), and the second term can be expressed as \(9{y^2} = {\left( {3y} \right)^2}\). Therefore, it does fit the desired form! Here, \(a = 2x\) and \(b = 3y\). So, we can factor it in the usual way.

$$\eqalign{

4{x^2} – 9{y^2} &= {\left( {2x} \right)^2} – {\left( {3y} \right)^2} \cr

& = \left( {2x – 3y} \right)\left( {2x + 3y} \right) \cr} $$

**Example 4**: Factor the binomial below.

$${x^4}\, -\, 16$$

This problem may seem quite different from the previous ones because we’re accustomed to dealing with variables raised to the power of \(2\). But here’s the key: the term \(x^4\) is actually a square in disguise since \({x^4} = {\left( {{x^2}} \right)^2}\). Similarly, we know that \(16 = {4^2}\). Therefore, it becomes clear that \(a = {x^2}\) and \(b = 4\).

$$\eqalign{

{x^4} – 16 &= {\left( {{x^2}} \right)^2} – {4^2} \cr

&= \left( {{x^2} – 4} \right)\left( {{x^2} + 4} \right) \cr} $$

I hope you’ll agree that we’re not finished yet. The binomial \({{x^2} – 4}\) is still a difference of squares, where \(a = x\) and \(b = 2\). Let’s go ahead and factor it further.

$$\eqalign{

{x^4} – 16 &= {\left( {{x^2}} \right)^2} – {4^2} \cr

&= \left( \color{red}{{{x^2} – 4}} \right)\left( {{x^2} + 4} \right) \cr

&= \left( \color{red}{{x – 2}} \right)\left(\color{red}{ {x + 2}} \right)\left( {{x^2} + 4} \right) \cr} $$