Answer
$x \in \{-1-i, -1+i, 3\}$
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, $a_{n}$.
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$f(x)=x^{3}-x^{2}-4x-6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm2,\pm3,\pm6,$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2,\pm3,\pm6$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 1 & -1& -4 & -6\\
& & 3& 6 & 6\\
& -- & -- & -- & --\\
& 1 & 2 & 2 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(x^{2} +2x+2)$
c. Solving 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, $x^2+2x+2$, $x= \frac{-2\pm\sqrt {2^2-4\times1 \times2}}{2\times 1}=-=\frac{-2\pm 2i}{2}=-1\pm i$
$x \in \{-1-i, -1+i, 3\}$
