Answer
$f(x)=(x+5)(x+4-i)(x+4+i)$
Zeros: $-5,\ \ -4+i,\ \ -4-i$
Work Step by Step
Degree 3: there are 3 complex zeros.
First, rational zero candidates: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 5,\pm 17,\pm 85}{\pm 1}$
Trying synthetic division, ... $x+5$
$\left.\begin{array}{l}
-5 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 13 &57 & 85 \\\hline
& -5 & -40 & -85 \\\hline
1 & 8 & 17 & |\ \ 0 \end{array}$
$f(x)=(x+5)(x^{2}+8x+17)$
... factoring the trinomial, we can't find two factors of $17$ that add up to $8$,
so we use the quadratic formula.
$x=\displaystyle \frac{-8\pm\sqrt{64-4(17)}}{2}=\frac{-8\pm\sqrt{-4}}{2}=\frac{-8\pm 2i}{2}=-4\pm i$
$f(x)=(x+5)(x+4-i)(x+4+i)$
Zeros: $-5,\ \ -4+i,\ \ -4-i$
