Answer
$x\displaystyle \in\left\{\frac{-1- \sqrt {7}i}{2}, \frac{-1 + \sqrt {7}i}{2}, 1, 3\right\}$ with a multiplicty of $1$
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, $a_{n}$.
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$f(x)=6x^4-18x^{3}+6x^{2}-30x+36$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36$
$q:\qquad \pm 1, \pm 2, \pm3, \pm6,$
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36 \pm \frac{1}{2},\pm\frac{1}{3}, \pm\frac{1}{6}, \pm\frac{2}{3}, \pm\frac{3}{2},\pm\frac{4}{3},\pm\frac{9}{2}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &6& -18 & 6 & -30 & 36\\
& & 6&-12 & -6 & -36\\
& -- & -- & -- & --\\
& 6&-12 & -6 & -36 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(6x^3-12x^{2} -6x-36)$
Try for $x=3$:
$\begin{array}{lllll}
\underline{3}| & 6 & -12 & -6 & -36\\
& & 18 & 18 & 36\\
& -- & -- & -- & --\\
& 6 & 6 & 12 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-1)(x-3)(6x^2+6x+12)$
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, $6x^2+6x+12$, $x= \frac{-6\pm\sqrt {6^2-4\times6 \times12}}{2\times 6}=-=\frac{-6\pm 6\sqrt {7}i}{12}=\frac{-1\pm \sqrt {7}i}{2}$
$x\displaystyle \in\left\{\frac{-1- \sqrt {7}i}{2}, \frac{-1 + \sqrt {7}i}{2}, 1, 3\right\}$ with a multiplicty of $1$
