Factorising Differences of Squares

The other special product you saw in the previous was the Product of Conjugates pattern. You used this to multiply two binomials that were conjugates. Here’s an example:

\(\begin{array}{c}\hfill (3x-4)(3x+4)\hfill \\ \hfill 9{x}^{2}-16\hfill \end{array}\)

Remember, when you multiply conjugate binomials, the middle terms of the product add to 0. All you have left is a binomial, the difference of squares.

Multiplying conjugates is the only way to get a binomial from the product of two binomials.

Product of Conjugates Pattern

If a and b are real numbers

\((a-b)(a+b)={a}^{2}-{b}^{2}\)

The product is called a difference of squares.

To factor, we will use the product pattern “in reverse” to factor the difference of squares. A difference of squares factors to a product of conjugates.

Difference of Squares Pattern

If a and b are real numbers,

This image shows the difference of two squares formula, a squared – b squared = (a – b)(a + b). Also, the squares are labeled, a squared and b squared. The difference is shown between the two terms. Finally, the factoring (a – b)(a + b) are labeled as conjugates.

Remember, “difference” refers to subtraction. So, to use this pattern you must make sure you have a binomial in which two squares are being subtracted.

Example: How to Factor Differences of Squares

Factor: \({x}^{2}-4\).

Solution

This table gives the steps for factoring x squared minus 4. The first step is identifying the pattern in the binomial including it is a difference. Also, the first and last terms are verified as squares.

The second step is writing the two terms as squares, x squared and 2 squared.

The second step is writing the two terms as squares, x squared and 2 squared. The third step is to write the factoring as a product of the conjugates (x – 2)(x + 2).

The last step is to check with multiplication.

Factor differences of squares.

\(\begin{array}{cccc}\mathbf{\text{Step 1.}}\;\text{Does the binomial fit the pattern?}\hfill & & & \hfill {a}^{2}-{b}^{2}\hfill \\ \phantom{\rule{2.5em}{0ex}}•\phantom{\rule{0.5em}{0ex}}\text{Is this a difference?}\hfill & & & \hfill \_\_\_\_-\_\_\_\_\hfill \\ \phantom{\rule{2.5em}{0ex}}•\phantom{\rule{0.5em}{0ex}}\text{Are the first and last terms perfect squares?}\hfill & & & \\ \mathbf{\text{Step 2.}}\;\text{Write them as squares.}\hfill & & & \hfill {(a)}^{2}-{(b)}^{2}\hfill \\ \mathbf{\text{Step 3.}}\;\text{Write the product of conjugates.}\hfill & & & \hfill (a-b)(a+b)\hfill \\ \mathbf{\text{Step 4.}}\;\text{Check by multiplying.}\hfill & & & \end{array}\)

It is important to remember that sums of squares do not factor into a product of binomials. There are no binomial factors that multiply together to get a sum of squares. After removing any GCF, the expression \({a}^{2}+{b}^{2}\) is prime!

Don’t forget that 1 is a perfect square. We’ll need to use that fact in the next example.

Example

Factor: \(64{y}^{2}-1\).

Solution


Is this a difference? Yes.
Are the first and last terms perfect squares?
Yes - write them as squares.
Factor as the product of conjugates.
Check by multiplying.
\((8y-1)(8y+1)\)
\(64{y}^{2}-1✓\)

Example

Factor: \(121{x}^{2}-49{y}^{2}\).

Solution

\(\begin{array}{cccc}& & & \hfill 121{x}^{2}-49{y}^{2}\hfill \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {(11x)}^{2}-{(7y)}^{2}\hfill \\ \text{Factor as the product of conjugates.}\hfill & & & \hfill (11x-7y)(11x+7y)\hfill \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}(11x-7y)(11x+7y)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}121{x}^{2}-49{y}^{2}\;✓\hfill & & & \end{array}\)

The binomial in the next example may look “backwards,” but it’s still the difference of squares.

Example

Factor: \(100-{h}^{2}\).

Solution

\(\begin{array}{cccc}& & & \hfill 100-{h}^{2}\hfill \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {(10)}^{2}-{(h)}^{2}\hfill \\ \text{Factor as the product of conjugates.}\hfill & & & \hfill (10-h)(10+h)\hfill \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}(10-h)(10+h)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}100-{h}^{2}\;✓\hfill & & & \end{array}\)

Be careful not to rewrite the original expression as \({h}^{2}-100\).

Factor \({h}^{2}-100\) on your own and then notice how the result differs from \((10-h)(10+h)\).

To completely factor the binomial in the next example, we’ll factor a difference of squares twice!

Example

Factor: \({x}^{4}-{y}^{4}\).

Solution

\(\begin{array}{cccc}& & & \hfill {x}^{4}-{y}^{4}\hfill \\ \text{Is this a difference of squares? Yes.}\hfill & & & \hfill {({x}^{2})}^{2}-{({y}^{2})}^{2}\hfill \\ \text{Factor it as the product of conjugates.}\hfill & & & \hfill ({x}^{2}-{y}^{2})({x}^{2}+{y}^{2})\hfill \\ \text{Notice the first binomial is also a difference of squares!}\hfill & & & \hfill ({(x)}^{2}-{(y)}^{2})({x}^{2}+{y}^{2})\hfill \\ \text{Factor it as the product of conjugates. The last}\hfill & & & \hfill (x-y)(x+y)({x}^{2}+{y}^{2})\hfill \\ \text{factor, the sum of squares, cannot be factored.}\hfill & & & \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}(x-y)(x+y)({x}^{2}+{y}^{2})\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}[(x-y)(x+y)]({x}^{2}+{y}^{2})\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}({x}^{2}-{y}^{2})({x}^{2}+{y}^{2})\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}{x}^{4}-{y}^{4}\;✓\hfill & & & \end{array}\)

As always, you should look for a common factor first whenever you have an expression to factor. Sometimes a common factor may “disguise” the difference of squares and you won’t recognize the perfect squares until you factor the GCF.

Example

Factor: \(8{x}^{2}y-18y\).

Solution

\(\begin{array}{cccc}& & & \hfill 8{x}^{2}y-98y\hfill \\ \text{Is there a GCF? Yes,}\;2y\text{—factor it out!}\hfill & & & \hfill 2y(4{x}^{2}-49)\hfill \\ \text{Is the binomial a difference of squares? Yes.}\hfill & & & \hfill 2y({(2x)}^{2}-{(7)}^{2})\hfill \\ \text{Factor as a product of conjugates.}\hfill & & & \hfill 2y(2x-7)(2x+7)\hfill \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}2y(2x-7)(2x+7)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}2y[(2x-7)(2x+7)]\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}2y(4{x}^{2}-49)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}8{x}^{2}y-98y\;✓\hfill & & & \end{array}\)

Example

Factor: \(6{x}^{2}+96\).

Solution

\(\begin{array}{cccc}& & & \hfill 6{x}^{2}+96\hfill \\ \text{Is there a GCF? Yes, 6—factor it out!}\hfill & & & \hfill 6({x}^{2}+16)\hfill \\ \text{Is the binomial a difference of squares? No, it}\hfill & & & \\ \text{is a sum of squares. Sums of squares do not factor!}\hfill & & & \\ \text{Check by multiplying.}\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}6({x}^{2}+16)\hfill & & & \\ \phantom{\rule{2.5em}{0ex}}6{x}^{2}+96\;✓\hfill & & & \end{array}\)

Factorising Differences of Squares