A binomial of the form \(a^2-b^2\) can be factored out as \(\left( {a + b} \right)\left( {a - b} \right)\), or \(\left( {a - b} \right)\left( {a + b} \right)\). The order of factors doesn't matter since multiplication is commutative. This is called the difference of two squares because we literally have a difference of two quantities that have been squared. In other words, we have the square of the first quantity \({a^2}\) subtracted by the square of the second quantity \({b^2}\), thus \({a^2} - {b^2}\).

Example 1: Factor the difference of two squares.

$${x^2} - 49$$

Observe that the first term is a square since the power of the variable \(x\) is \(2\). The second term is also a square since it can be expressed as a square of an integer. That is, \(49 = {7^2}\). Now, we can rewrite 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 we can write it as a square of a positive integer, that is, \(121 = {11^2}\). The criteria are met so we can proceed to factor this out. Note that \(a=y\) 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 may not look like a difference of two squares. But notice that the first term can be written as \(4{x^2} = {\left( {2x} \right)^2}\) while the second term can be expressed as \(9{y^2} = {\left( {3y} \right)^2}\). So yes, it is in the form that we want! Here, \(a = 2x\) and \(b = 3y\). Therefore, we can factor it as usual.

$$\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 looks a whole lot different than the previous ones because we are used to seeing variables with power of \(2\). Here's the deal. The term \(x^4\) is actually a square term in disguise since \({x^4} = {\left( {{x^2}} \right)^2}\). We know that \(16 = {4^2}\). It should be clear now 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 agree with me that we are not done yet. The binomial \({{x^2} - 4}\) is still a case of difference of squares where \(a=x\) and \(b=2\). Let's factor it out 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} $$