When we multiply together the two binomials $(x-y)$ and $(x+y)$, we obtain the product $x^2-y^2.$ Usually when we multiply two binomials we obtain a trinomial, but in this case, when we FOIL, the outer and inner terms cancel. When we recall this fact in the reverse direction, it is called the difference of squares formula:

 $x^2-y^2=(x-y)(x+y)$

Another way to think about this formula is to consider trying to solve the equation $x^2=a^2$ for $x$. Taking the square root of both sides of the equation gives the answer $x = \pm a$. 

But $x^2 = a^2$ can also be solved by rewriting the equation as $x^2-a^2=0$ and factoring the difference of squares. From this, you would obtain:

$(x-a)(x+a)=0$ 

Thus  there are two solutions, where $x-a=0$ ($x=a$) and where $x+a=0$, ($x=-a$); the same as above. Using the difference of squares is just another way to think about solving the equation.

Example 1

Suppose you were trying to factor the quadratic expression: 

$x^2-16$ 

If you recognize the first term as the square of $x$ and the term after the minus sign as the square of $4$, you can then factor the expression as: 

$(x-4)(x+4)$

Example 2

Suppose you were trying to solve the equation: 

$16x^4=9$ 

We could write this as: 

$16x^4-9=0$ 

If we recognize the first term as the square of $4x^2$ and the term after the minus sign as the square of $3$, we can rewrite the equation as:

$(4x^2-3)(4x^2+3)=0$ 

This means that either $4x^2-3=0$ or $4x^2+3=0$. This latter equation has no solutions, since $4x^2$ is always greater than or equal to $0.$ However, the first equation $4x^2-3=0$ can be factored again as the difference of squares, if we consider $3$ as the square of $\sqrt3$. 

Thus we have $(2x-\sqrt3)(2x+\sqrt3)=0.$ 

Our two solutions occur where $2x-\sqrt3=0$ (which is at $x=\frac{\sqrt3}{2}$), and where $2x+\sqrt3=0$ (which is at $x=-\frac{\sqrt3}{2}$).