Answer
$x\displaystyle \in\{-2, -1, 3 - \sqrt {13}, 3 + \sqrt {13} \}$
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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$f(x)=x^4-3x^{3}-20x^{2}-24x-8$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 4, \pm 8$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 4, \pm 8$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & -3 & -20 & -24 & -8\\
& & -1 & 4 & 16& 8\\
& -- & -- & -- & --\\
& 1 & -4 & -16 & -8 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^3-4x^{2} -16x-8)$,
to solve for quadrinomial,
Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| & 1 & -4 & -16 & -8\\
& & -2 & 12 & 8\\
& -- & -- & -- & --\\
& 1 & -6 & -4 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+1)(x+2)(x^2-6x-4)$
c. To solve for the trinomial, using the quadratic function, $x=\frac {-b \pm \sqrt {b^2-4ac}}{2a}$.
In this case, $a=1,b=-6,c=-4$.
Therefore, $x=\frac {6 \pm \sqrt {(-6)^2-4\times 1\times (-4)}}{2\times1}$,
$x=\frac {6 \pm 2\sqrt {13}}{2}$,
$x=3 \pm \sqrt {13}$
The zeros of f satisfy $f(x)=0$
$(x+1)(x+2)(x-3 \pm \sqrt {13})=0$
$x\displaystyle \in\{-2, -1, 3 - \sqrt {13}, 3 + \sqrt {13} \}$
