Answer
$x\displaystyle \in\{-3, -2-\sqrt 2i,-2+\sqrt 2i \}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^{3}+7x^{2}+18x+18$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3, \pm6,\pm9, \pm18$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm2, \pm3, \pm6, \pm9, \pm18 $
b. Try for $x=-3:$
$\begin{array}{lllll}
\underline{-3}| & 1 & 7 & 18 & 18\\
& & -3 & -12 & -18\\
& -- & -- & -- & --\\
& 1 & 4 & 6 & |\underline{0}
\end{array}$
$-3$ is a zero,
$f(x)=(x+3)(x^{2} +4x+6)$
c. Solve for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
in this case, $x^2+4x+6$, $x=\frac{-4 \pm \sqrt {4^2-4\times 1\times 6}}{2}=\frac{-4\pm 2\sqrt {2}i}{2}=-2\pm \sqrt 2i$
$x\displaystyle \in\{-3, -2-\sqrt 2i,-2+\sqrt 2i \}$
