Answer
$x \in \left\{-2, -i, i, \frac{1}{2}\right\}$
$f(x)=(x+2)(x+i)(x-i)(2x-1)$
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, $a_{n}$.
------------------------
$f(x)=2x^4+3x^{3}+3x-2$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm\frac{1}{2}$
b. Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| & 2 & 3 & 0 & 3& -2\\
& & -4 & 2 & -4& 2\\
& -- & -- & -- & --\\
& 2 & -1 & 2 & -1& |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+2)(2x^3-x^{2} +2x-1)$
c. Factorize the Quadrinomial $x^2(2x-1)+1(2x-1)$...
$(x^2+1)(2x-1)$
$f(x)=(x+2)(x+i)(x-i)(2x-1)$
The zeros of f satisfy $f(x)=0$
$(x+2)(x^2+1)(2x-1)=0$
$x \in \left\{-2, -i, i, \frac{1}{2}\right\}$
