Answer
$x\displaystyle \in\{-2, -3i, 3i, 1\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ with integer coefficients, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$.
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$f(x)=x^4+x^{3}+7x^2+9x-18$
a. candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 6, \pm9, \pm18$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 6, \pm9, \pm18$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 1 & 1 & 7 & 9& -18\\
& & 1 & 2 & 9& 18\\
& -- & -- & -- & --\\
& 1 & 2 & 9& 18 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^3+2x^{2} +9x+18)$
Try for $x=-2$.
$\begin{array}{lllll}
\underline{-2}| & 1 & 2 & 9& 18\\
& & -2 & 0& -18\\
& -- & -- & -- & --\\
& 1 & 0& 9 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x-1)(x+2)(x^2+9)$,
c. $x^2+9=0$, $x=\pm 3i$.
$x\displaystyle \in\{-2, -3i, 3i, 1\}$
