Factoring Using Perfect Square Trinomials

Example 1

\(\begin{align*}x^2 + x + 0.25 = 0\end{align*}\)

\(\begin{align*}x^2 + x + 0.25 = 0\end{align*}\)

Rewrite: The equation is in the correct form already.

Factor: Rewrite \(\begin{align*}x^2 + x + 0.25 =0\end{align*}\) as \(\begin{align*}x^2 + 2 \cdot (0.5)x + ( 0.5)^2\end{align*}\).

We recognize this as a perfect square. This factors as \(\begin{align*}(x +0.5)^2 = 0 \end{align*}\) or \(\begin{align*}(x +0.5)(x +0.5) = 0\end{align*}\)

Set the factor equal to zero:

\(\begin{align*}x +0.25 = 0 \end{align*}\)

Solve:

\(\begin{align*}\underline{\underline{x = -0.5}}\end{align*}\)

Check: Substitute the solution back into the original equation.

\(\begin{align*}(-0.5)^2 + -0.5 + 0.25&= \quad \quad \text{Substitute in -0.5}\\ 0.25+ -0.5 + 0.25&= \quad \quad \text{Simplify.}\\ 0.5-0.5&=0 \quad \quad \text{Checks out.} \end{align*}\)

Example 2

\(\begin{align*}x^2 - 81 = 0\end{align*}\)

\(\begin{align*}x^2 - 81 = 0\end{align*}\)

Rewrite: this is not necessary since the equation is in the correct form already

Factor: Rewrite \(\begin{align*}x^2 - 81\end{align*}\) as \(\begin{align*}x^2 - 9^2\end{align*}\).

We recognize this as a difference of squares. This factors as \(\begin{align*}(x - 9)(x + 9) = 0\end{align*}\).

Set each factor equal to zero:

\(\begin{align*}x - 9 = 0 && \text{or} && x + 9 = 0\end{align*}\)

Solve:

\(\begin{align*}\underline{\underline{x = 9}} && \text{or} && \underline{\underline{x = - 9}}\end{align*}\)

Check: Substitute each solution back into the original equation.

\(\begin{align*}& x = 9 && 9^2 - 81 = 81-81 = 0 && \text{checks out}\\ & x = -9 && (-9)^2 - 81 = 81 - 81 = 0 && \text{checks out}\end{align*}\)

c) \(\begin{align*}x^2 + 20x + 100 =0\end{align*}\)

Rewrite: this is not necessary since the equation is in the correct form already

Factor: Rewrite \(\begin{align*}x^2 + 20x + 100\end{align*}\) as \(\begin{align*}x^2 + 2 \cdot 10 \cdot x + 10^2\end{align*}\).

We recognize this as a perfect square. This factors as \(\begin{align*}(x + 10)^2 =0\end{align*}\) or \(\begin{align*}(x + 10)(x + 10)=0\end{align*}\)

Set each factor equal to zero:

\(\begin{align*}x + 10 =0 && \text{or} && x + 10 = 0\end{align*}\)

Solve:

\(\begin{align*}\underline{\underline{x = -10}} && \text{or} && \underline{\underline{x = -10}} \quad \quad \text{This is a double root.}\end{align*}\)

Check: Substitute each solution back into the original equation.

\(\begin{align*}x = 10 && (-10)^2 + 20(-10) + 100 = 100 -200 + 100 =0 && \text{checks out}\end{align*}\)