Factoring Using Perfect Square Trinomials

1. Factor the following perfect square trinomials:

a) \(\begin{align*}x^2 + 8x +16\end{align*}\)

The first step is to recognize that this expression is a perfect square trinomial.

First, we can see that the first term and the last term are perfect squares. We can rewrite \(\begin{align*}x^2 + 8x + 16\end{align*}\) as \(\begin{align*}x^2 + 8x + 4^2\end{align*}\).

Next, we check that the middle term is twice the product of the square roots of the first and the last terms. This is true also since we can rewrite \(\begin{align*}x^2 + 8x + 16\end{align*}\) as \(\begin{align*}x^2 + 2 \cdot 4 \cdot x + 4^2\end{align*}\).

This means we can factor \(\begin{align*}x^2 + 8x + 16\end{align*}\) as \(\begin{align*}(x + 4)^2\end{align*}\). We can check to see if this is correct by multiplying \(\begin{align*}(x + 4)^2 = (x + 4)(x + 4)\end{align*}\) :

\(\begin{align*}& \quad \quad \quad x + 4\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;x + 4}\\ & \quad \quad \ 4x + 16\\ & \underline{x^2 + 4x\;\;\;\;\;\;\;\;}\\ & x^2 + 8x + 16\end{align*}\)

The answer checks out.

Note: We could factor this trinomial without recognizing it as a perfect square. We know that a trinomial factors as a product of two binomials:

\(\begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}\)

We need to find two numbers that multiply to 16 and add to 8. We can write 16 as the following products:

\(\begin{align*}& 16 = 1 \cdot 16 && \text{and} && 1 + 16 = 17\\ & 16 = 2 \cdot 8 && \text{and} && 2 + 8 = 10\\ & 16 = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \qquad These \ are \ the \ correct \ numbers\end{align*}\)

So we can factor \(\begin{align*}x^2 + 8x + 16\end{align*}\) as \(\begin{align*}(x + 4)(x + 4)\end{align*}\), which is the same as \(\begin{align*}(x + 4)^2\end{align*}\).

Once again, you can factor perfect square trinomials the normal way, but recognizing them as perfect squares gives you a useful shortcut.

b) \(\begin{align*}x^2 - 4x + 4\end{align*}\)

Rewrite \(\begin{align*}x^2 + 4x + 4\end{align*}\) as \(\begin{align*}x^2 + 2 \cdot (-2) \cdot x + (-2)^2\end{align*}\).

We notice that this is a perfect square trinomial, so we can factor it as \(\begin{align*}(x - 2)^2\end{align*}\).

c) \(\begin{align*}x^2 + 14x +49\end{align*}\)

Rewrite \(\begin{align*}x^2 + 14x + 49\end{align*}\) as \(\begin{align*}x^2 + 2 \cdot 7 \cdot x + 7^2\end{align*}\).

We notice that this is a perfect square trinomial, so we can factor it as \(\begin{align*}(x + 7)^2\end{align*}\).

2. Factor the following perfect square trinomials:

a) \(\begin{align*}4x^2 + 20x + 25\end{align*}\)

Rewrite \(\begin{align*}4x^2 + 20x + 25\end{align*}\) as \(\begin{align*}(2x)^2 + 2 \cdot 5 \cdot (2x) + 5^2\end{align*}\).

We notice that this is a perfect square trinomial and we can factor it as \(\begin{align*}(2x + 5)^2\end{align*}\).

b) \(\begin{align*}9x^2 - 24x + 16\end{align*}\)

Rewrite \(\begin{align*}9x^2 - 24x + 16\end{align*}\) as \(\begin{align*}(3x)^2 + 2 \cdot (-4) \cdot (3x) + (-4)^2\end{align*}\).

We notice that this is a perfect square trinomial and we can factor it as \(\begin{align*}(3x - 4)^2\end{align*}\).

We can check to see if this is correct by multiplying \(\begin{align*}(3x - 4)^2 = (3x - 4)(3x - 4)\end{align*}\):

\(\begin{align*}& \quad \quad \quad 3x - 4\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;3x - 4\;\;}\\ & \quad \ -12x + 16\\ & \underline{9x^2 - 12x\;\;\;\;\;\;\;\;}\\ & 9x^2 - 24x + 16\end{align*}\)

The answer checks out.

c) \(\begin{align*}x^2 + 2xy + y^2\end{align*}\)

\(\begin{align*}x^2 + 2xy + y^2\end{align*}\)

We notice that this is a perfect square trinomial and we can factor it as \(\begin{align*}(x + y)^2\end{align*}\).