Answer
$f(x)=2(x-5)(x-\displaystyle \frac{1}{2})(x+3-i )(x+3+3i )$
Zeros: $\ \ 5,\displaystyle \ \frac{1}{2},\ -3-2i,\ -3+2i$
Work Step by Step
Degree $4$: there are $4$ complex zeros.
First, rational zero candidates: $\displaystyle \frac{p}{q}=$
$\displaystyle \frac{\pm 1,\pm 2, \pm 5, \pm 13, \pm 65 }{\pm 1,\pm 2}$
Trying synthetic division, ... $x-5$
$\left.\begin{array}{l}
5 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 1 &-35 & -113 & 65 \\\hline
& 10 & 55 & 100 & -65 \\\hline
2 & 11 & 20 & -13 & |\ \ 0 \end{array}$
$f(x)=(x-5)(2x^{3}+11x^{2}+20x-13)$
... factor in pairs not obvious... try synthetic division $x-\displaystyle \frac{1}{2}$
$\left.\begin{array}{l}
1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
2 & 11 & 20 & -13 \\\hline
& 1 & 6 & +13 \\\hline
2 & 12 & 26 & |\ \ 0 \end{array}$
$f(x)=(x-5)(x-\displaystyle \frac{1}{2})(2x^{2}+12x+26)$
$=(x-5)(x-\displaystyle \frac{1}{3})\cdot 2(x^{2}+6x+13)$
use the quadratic formula on $x^{2}+6x+13=0$
$x=\displaystyle \frac{-6\pm\sqrt{36-4(13)}}{2 }=\frac{-6\pm\sqrt{-16}}{2}=\frac{-6\pm 4i}{2}=-3\pm 2i$
$f(x)=2(x-5)(x-\displaystyle \frac{1}{2})(x+3-i )(x+3+3i )$
Zeros: $5,\displaystyle \ \frac{1}{2},\ -3-2i,\ -3+2i$
