## Perfect Square Trinomials

Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form.

$\begin{array}{ccc}\hfill \text{ }{\left(x+5\right)}^{2}& =& \text{ }{x}^{2}+10x+25\hfill \\ \hfill {\left(x - 3\right)}^{2}& =& \text{ }{x}^{2}-6x+9\hfill \\ \hfill {\left(4x - 1\right)}^{2}& =& 4{x}^{2}-8x+1\hfill \end{array}$

Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial.

### A General Note: Perfect Square Trinomials

When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared.

${\left(x+a\right)}^{2}=\left(x+a\right)\left(x+a\right)={x}^{2}+2ax+{a}^{2}$

### How To: Given a binomial, square it using the formula for perfect square trinomials.

1. Square the first term of the binomial.
2. Square the last term of the binomial.
3. For the middle term of the trinomial, double the product of the two terms.

### Example 6: Expanding Perfect Squares

${\left(3x - 8\right)}^{2}$.

### Solution

Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms.

${\left(3x\right)}^{2}-2\left(3x\right)\left(8\right)+{\left(-8\right)}^{2}$

$9{x}^{2}-48x+64$.

### Try It 6

${\left(4x - 1\right)}^{2}$.

Solution

## Difference of Squares

Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply $\left(x+1\right)\left(x - 1\right)$ using the FOIL method.

$\begin{array}{ccc}\hfill \left(x+1\right)\left(x - 1\right)& =& {x}^{2}-x+x - 1\hfill \\ & =& {x}^{2}-1\hfill \end{array}$

The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.

$\begin{array}{ccc}\hfill \left(x+5\right)\left(x - 5\right)& =& {x}^{2}-25\hfill \\ \hfill \left(x+11\right)\left(x - 11\right)& =& {x}^{2}-121\hfill \\ \hfill \left(2x+3\right)\left(2x - 3\right)& =& 4{x}^{2}-9\hfill \end{array}$

Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.

### Is there a special form for the sum of squares?

No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares.

### A General Note: Difference of Squares

When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term.

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

### How To: Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares.

1. Square the first term of the binomials.
2. Square the last term of the binomials.
3. Subtract the square of the last term from the square of the first term.

### Example 7: Multiplying Binomials Resulting in a Difference of Squares

Multiply $\left(9x+4\right)\left(9x - 4\right)$.

### Solution

Square the first term to get ${\left(9x\right)}^{2}=81{x}^{2}$. Square the last term to get ${4}^{2}=16$. Subtract the square of the last term from the square of the first term to find the product of $81{x}^{2}-16$.

### Try It 7

Multiply $\left(2x+7\right)\left(2x - 7\right)$.

Solution