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