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A difference of two squares is an expression in the form $x^2 - y^2$. It expands to the expression $(x + y)(x - y)$. When multiplying the two terms $(x + y)$ and $(x - y)$, the coefficient of the term with both $x$ and $y$ is equal to $0.$ That leaves us with only the $x^2$ term and the $y^2$ term.

Here are some examples:

Now let's try going backwards:

You could've used FOIL or tables!

Difference of Squares

There's only one big thing you need to remember for this topic, and it's this equation:

Notice that the left side of the equation above is literally the difference (i.e. subtraction) of two perfect squares.

Every time you see an expression in which one perfect square is being subtracted from another--as is depicted on the left side of the equation above--you can expand that expression into the form that you see displayed on the equation's right side.

Let's look at some examples that apply this equation to differences of perfect squares.

Yes! Both $9x^2$ and $16y^2$ are perfect squares, and one of those terms is being subtracted from the other.

Now that we've determined the square roots of both terms in our expression, we're ready to apply the equation that we previously learned:

And we're done!

Let's take a look at another example.

If you want to review perfect squares, check out this lesson!

You can also use the difference of squares equation to solve problems that do not contain variables.

(Video) The Difference of Squares

by mathman1024

This by mathman1024 that shows how to employ the difference of squares method in factoring binomials.

$\large{\text{Summary}}$

This method makes use of the difference of squares equation, which states the following:

$$a^2-b^2=(a+b)(a-b)$$

The first binomial that our instructor factors is:

$$x^2-25$$

However, you might notice that we can rewrite this binomial like so:

$$(x)^2-(5)^2$$

As our instructor points out, we can let $x$ be our $a$ in our difference of squares equation, and we can let $5$ be our $b$ in our difference of squares equation. Then, we just have to substitute those terms into the right-hand side of our equation:

$$(x+5)(x-5)$$

He then goes over some additional problems, and some of them are more complex than others. See how many you can figure out on your own, and then watch the video above to see if you were correct.

1. $x^2-49$
2. $4x^2-9$
3. $x^2-\frac{121}{64}$
4. $x^{10}-16$
5. $49x^4-81y^{12}$
6. $100x^7-25x^3y^4$ (Hint: do the terms in this binomial have a GCF?)
7. $x^4-81$ (Hint: can either of this binomial's factors be factored again?)