Answer
$x\displaystyle \in\left\{\frac{1- \sqrt 5i}{2}, \frac{1+ \sqrt 5i}{2}, 3 \right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $$ 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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$f(x)=2x^{3}-8x^2+9x-9$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}$
b. Try for $x=3:$
$\begin{array}{lllll}
\underline{3}| & 2 & -8 & 9 & -9\\
& & 6 & -6 & 9\\
& -- & -- & -- & --\\
& 2 & -2 & 3 & |\underline{0}
\end{array}$
$3$ is a zero,
$f(x)=(x-3)(2x^{2} -2x+3)$
c. Solve 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, $2x^2-2x+3$, $x=\frac{2 \pm \sqrt {(-2)^2-4\times 2\times 3}}{2\times 2}=\frac{2\pm 2\sqrt {5}i}{4}=\frac{1\pm \sqrt 5i}{2}$
$x\displaystyle \in\left\{\frac{1- \sqrt 5i}{2}, \frac{1+ \sqrt 5i}{2}, 3 \right\}$
