Answer
$x\displaystyle \in\{ -2i, 2i, 3\}$
The zero $3$ has multiplicity $2$
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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$f(x)=x^4-6x^{3}+13x^2-24x+36$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad\pm 1, \pm 2, \pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 1 & -6 & 13 & -24& 36\\
& & 3 & -9 & 12& -36\\
& -- & -- & -- & --\\
& 1 & -3 & 4& -12 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(x^3-3x^{2} +4x-12)$
Try for $x=3$.
$\begin{array}{lllll}
\underline{3}| & 1 & -3 & 4& -12\\
& & 3 & 0& 12\\
& -- & -- & -- & --\\
& 1 & 0& 4 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)^2(x^2+4)$,
c. $x^2+4=0$, $x=\pm 2i$.
$x\displaystyle \in\{ -2i, 2i, 3\}$
The zero $3$ has multiplicity $2$
