Answer
$x\displaystyle \in\left\{-1- \sqrt 2i, -1+ \sqrt 2i, -\frac{3}{2} \right\}$
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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$f(x)=2x^{3}+7x^2+12x+9$
a. candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$
b. Try for $x=-\frac{3}{2}:$
$\begin{array}{lllll}
\underline{-\frac{3}{2}}| & 2 & 7 & 12 & 9\\
& & -3 & -6 & -9\\
& -- & -- & -- & --\\
& 2 & 4 & 6 & |\underline{0}
\end{array}$
$-\frac{3}{2}$ is a zero,
$f(x)=(x+\frac{3}{2})(2x^{2} +4x+6)$
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, $2x^2+4x+6$, $x=\frac{-4 \pm \sqrt {4^2-4\times 2\times 6}}{2\times 2}=\frac{-4\pm 4\sqrt {2}i}{4}=-1\pm \sqrt 2i$
$x\displaystyle \in\left\{-1- \sqrt 2i, -1+ \sqrt 2i, -\frac{3}{2} \right\}$
