## Precalculus (6th Edition)

$\left\{-2, 4\right\}$
RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The given trinomial has $b=2$ and $c=-8$. Note that $-8=-4(2)$ and $2= -4+2$. This means that $d=-4$ and $e=2$ Thus, the factored form of the trinomial is: $[x+(-4)](x+2) = (x-4)(x+2)$ The given equation maybe written as: $(x-4)(x+2)=0$ Use the Zero-Factor Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &x-4 = 0 &\text{ or } &x+2=0 \\&x=4 &\text{ or } &x=-2 \end{array}$ Thus, the solution set is $\left\{-2, 4\right\}$.