Answer
$f(x)=(x-1)(x+3)(x+5i )(x-5i )$
Zeros: $1,\ -3,\ 5i,\ -5i$
Work Step by Step
Degree $4$: there are $4$ complex zeros.
First, rational zero candidates: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 3,\pm 5,\pm 15,\pm 25,\pm 75}{\pm 1}$
Trying synthetic division, ... $x-1$
$\left.\begin{array}{l}
1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 2 &22 & 50 & -75 \\\hline
& 1 & 3 & 25 & +75 \\\hline
1 & 3 & 25 & 75 & |\ \ 0 \end{array}$
$f(x)=(x-1)(x^{3}+3x^{2}+25x+75)$
... factor in pairs ...
$x^{3}+3x^{2}+25x+75=x^{2}(x+3)+25(x+3)=(x+3)(x^{2}+25)$
$x^{2}+25=0\Rightarrow x=\pm 5i$
$f(x)=(x-1)(x+3)(x+5i )(x-5i )$
Zeros: $1,\ -3,\ 5i,\ -5i$
