Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{-2, 9\right\}}$
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 trinomial in the equation has $b=-7$ and $c=-18$. Note that $-18=-9(2)$ and $-7= -9+2$. This means that $d=-9$ and $e=2$ Thus, the factored form of the trinomial is: $[w+(-9)](w+2)=(w-9)(w+2)$ The given equation maybe written as: $(w-9)(w+2)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{array}{ccc} &w-9 = 0 &\text{ or } &w+2=0 \\&w=9&\text{ or } &w=-2 \end{array} Thus, the solution set is $\color{blue}{\left\{-2, 9\right\}}$.