Answer
$f(x)=(x-1)(x-1)(x-2i)(x+2i)$
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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$f(x)=x^4-2x^{3}+5x^{2}-8x+4$
a. Candidates for zeros, $ \frac{p}{q}:$
$p:\qquad \pm 1, \pm2,\pm4,$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2,\pm4$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &1& -2& 5 & -8 & 4\\
& & 1 &-1 & 4 & -4\\
& -- & -- & -- & --\\
& 1&-1 & 4& -4 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^3-x^{2} +4x-4)$
c. Factorising the quadrinomial, $x^3-x^2+4x-4$,
$x^3-x^2+4x-4=x^2(x-1)+4(x-1)=(x^2+4)(x-1)$,
thus, $f(x)=(x-1)(x-1)(x^2+4)$.
zeros of $f$ satisfy:
$(x-1)^2(x^2+4)=0$
$x-1=0, x=1$ or $x^2+4=0, x=\pm2i$,
$f(x)=(x-1)(x-1)(x-2i)(x+2i)$
$x\displaystyle \in\{-2i, 2i, 1\}$
