A perfect square trinomial is a special type of trinomial because it can be expressed as the square of a binomial. The trick is to recognize the pattern.

These are the two versions of the perfect square trinomial. We can use the following patterns to write the trinomial as a square of a binomial.

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

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

Notice that the first and last terms are square terms. The first term is \(a^2\), while the last term is \(b^2\). In addition, the middle term is just twice ( two times) the product of the terms that are being squared, namely \(a\) and \(b\). That is, \(2ab\). For now, just disregard the sign of \(2ab\). If these two conditions are met, we have a perfect square trinomial which can be rewritten as the square of a binomal.

Example 1: Factor the trinomial below.

$${x^2} + 8x + 16$$

The first term \(x^2\) is a square. The term that is being squared is \(x\). That means \(a=x\). The last term \(16\) is also a square. The term that is being squared is \(4\). That means \(b=4\).

Now for the middle term, is it equal to \(2ab\)? Let's check! From above, since \(a=x\) and \(b=4\), we have \(2ab = 2\left( x \right)\left( 4 \right) = 8x\). Yes! This is a perfect square trinomial. Since the middle term is positive, we will use the first version of the formula.

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

Since \(a=x\) and \(b=4\), we have

$$\eqalign{

{x^2} + 8x + 16 &= {\left( x \right)^2} + 2\left( x \right)\left( 4 \right) + {\left( 4 \right)^2} \cr

& = {\left( {x + 4} \right)^2} \cr

& = \left( {x + 4} \right)\left( {x + 4} \right) \cr} $$

Example 2: Factor the trinomial below.

$${x^2} - 16x + 64$$

Observe that both the first and last terms are square terms where \(a=x\) and \(b=8\). Disregarding the sign for now, the middle term is simply twice the product of \(a\) and \(b\), that is \(2ab=2(x)(8)=16x\). Because the middle term is negative, we will use the second version of the formula.

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

$$\eqalign{

{x^2} - 16x + 64 &= {\left( x \right)^2} - 2\left( x \right)\left( 8 \right) + {\left( 8 \right)^2} \cr

& = {\left( {x - 8} \right)^2} \cr

& = \left( {x - 8} \right)\left( {x - 8} \right) \cr} $$

Example 3: Factor the trinomial below.

$$9{x^2} - 12x + 4$$

By inspection, the first term \(9x^2\) is a square term because \(9x^2=(3x)^2\). The last term is also a square since \(4=2^2\). So we have \(a=3x\) and \(b=2\). That means \(2ab=2(3x)(2)=12x\) which matches the middle term (disregard the sign). Since the two conditions are met, this is a perfect square trinomial. We will use the second version because the middle term is negative.

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

From above, we have \(a=3x\) and \(b=2\). This factors out as

$$\eqalign{

9{x^2} - 12x + 4 &= {\left( {3x} \right)^2} - 2\left( {3x} \right)\left( 2 \right) + {\left( 2 \right)^2} \cr

& = {\left( {3x - 2} \right)^2} \cr

& = \left( {3x - 2} \right)\left( {3x - 2} \right) \cr} $$