Answer
$c=-2$ is a zero.
$P(x)=(x+2)(x-3)(x+3),$
$-3$ and $3$ are also zeros of P.
Work Step by Step
$\left.\begin{array}{l}
-2 \ \lfloor \\ \\ \\ \end{array}\right.\begin{array}{rrrrr}
1 & 2 & -9 & -18 \\\hline
& -2 & 0 & 18 \\\hline
1 & 0 & -9 &|\ \ \ 0 \end{array}$
$Q(x)=x^{2}-9,\quad R(x)=0$
By the remainder theorem, $P(-2)=0$, so $c=-2$ is a zero.
$P(x)=D(x)\cdot Q(x)+R(x)=(x+2)(x^{2}-9),$
Factoring the second set of parentheses (as a difference of squares), we get
$P(x)=(x+2)(x-3)(x+3),$
$-3$ and $3$ are also zeros of P
