Example 8.7.1 Solving Using Factoring
Solve the quadratic equations using factoring.
\(x^22x15=0\)
\(4x^240x=96\)
\(6x^2+x12=0\)
\((x3)(x+2)=14\)
\(x^364x=0\)

Use factor pairs.
\begin{align*} x^22x15\amp=0\\ (x5)(x+3)\amp=0 \end{align*}\begin{align*} x5\amp=0 \amp\text{ or }\amp\amp x+3\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=3 \end{align*}So the solution set is \(\{5,3\}\text{.}\)

Start by putting the equation in standard form and factoring out the greatest common factor.
\begin{align*} 4x^240x\amp=96\\ 4x^240x+96\amp=0\\ 4\left(x^210x+24\right)\amp=0\\ 4(x6)(x4)\amp=0 \end{align*}\begin{align*} x6\amp=0 \amp\text{ or }\amp\amp x4\amp=0\\ x\amp=6 \amp\text{ or }\amp\amp x\amp=4 \end{align*}So the solution set is \(\{4,6\}\text{.}\)

Use the AC method.
\begin{align*} 6x^2+x12\amp=0\\ \end{align*}Note that \(a\cdot c=72\) and that \(\highlight{9\cdot8}=72\) and \(\highlight{98}=1\)
\begin{align*} 6x^2\substitute{+9x8x}12\amp=0\\ \left(6x^2+9x\right)+\left(8x12\right)\amp=0\\ \highlight{3x}\left(2x+3\right)\mathbin{\highlight{4}}\left(2x+3\right)\amp=0\\ \left(2x+3\right)\highlight{\left(3x4\right)}\amp=0 \end{align*}\begin{align*} 2x+3\amp=0 \amp\text{ or }\amp\amp \highlight{3x4}\amp=0\\ x\amp=\frac{3}{2} \amp\text{ or }\amp\amp x\amp=\highlight{\frac{4}{3}} \end{align*}So the solution set is \(\left\{\frac{3}{2},\frac{4}{3}\right\}\text{.}\)

Start by putting the equation in standard form.
\begin{align*} (x3)(x+2)\amp=14\\ x^2x6\amp=14\\ x^2x20\amp=0\\ (x5)(x+4)\amp=0 \end{align*}\begin{align*} x5\amp=0 \amp\text{ or }\amp\amp x+4\amp=0\\ x\amp=5 \amp\text{ or }\amp\amp x\amp=4 \end{align*}So the solution set is \(\{5,4\}\text{.}\)

Even though this equation has a power higher than \(2\text{,}\) we can still find all of its solutions by following the algorithm. Start by factoring out the greatest common factor.
\begin{align*} x^364x\amp=0\\ x\left(x^264\right)\amp=0\\ x(x8)(x+8)\amp=0 \end{align*}\begin{align*} x\amp=0 \amp\text{ or }\amp\amp x8\amp=0 \amp\text{ or }\amp\amp x+8\amp=0\\ x\amp=0 \amp\text{ or }\amp\amp x\amp=8 \amp\text{ or }\amp\amp x\amp=8 \end{align*}So the solution set is \(\{0,8,8\}\text{.}\)