Answer
$\left\{-4,2\right\}$
Work Step by Step
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 $e=-2$
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 $\left\{-4, 2\right\}$.
