Answer
$x\displaystyle \in\{ -3i,3i, 1\}$
The zero $1$ has multiplicity $3$.
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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$f(x)=x^5-3x^4+12x^3-28x^2+27x-9$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1,\pm3,\pm9$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad\pm 1, \pm3,\pm9$
Try for $x=1$.
$\begin{array}{lllll}
\underline {1}| & 1& -3 & 12 & -28&27& -9\\
& & 1&-2 & 10&-18& 9\\
& -- & -- & -- & --\\
& 1&-2 &10& -18& 9 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^4-2x^3+10x^2-18x+9)$,
Try for $x=1$.
$\begin{array}{lllll}
\underline {1}| & 1 & -2 & 10&-18& 9\\
& & 1 & -1&9& -9\\
& -- & -- & -- & --\\
& 1 &-1& 9& -9 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)^2(x^3-x^2+9x-9)$,
c. Factor the quadrinomial:
$x^3-x^2+9x-9=x^2(x-1)+9(x-1)=(x^2+9)(x-1)=(x^2+9)(x-1)$.
Solve the equation $(x-1)^3(x^2+9)=0$
$x\displaystyle \in\{ -3 i,3i, 1\}$
The zero $1$ has multiplicity $3$.
