Answer
$f(x)=(x+4)(x-\displaystyle \frac{1}{3})(x-2+3i )(x-2-3i )$
Zeros: $-4,\displaystyle \ \frac{1}{3},\ 2-3i,\ 2+3i$
Work Step by Step
Degree $4$: there are $4$ complex zeros.
First, rational zero candidates: $\displaystyle \frac{p}{q}=$
$\displaystyle \frac{\pm 1,\pm 2, \pm 4, \pm 13, \pm 26, \pm 52 }{\pm 1}$
Trying synthetic division, ... $x+4$
$\left.\begin{array}{l}
-4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 & - 1 &-9 & 159 & -52 \\\hline
& -12 & 52 & -172 & +52 \\\hline
3 & -13 & 43 & -13 & |\ \ 0 \end{array}$
$f(x)=(x+4)(3x^{3}-13x^{2}+43x-13)$
... factor in pairs not obvious... try synthetic division $x-\displaystyle \frac{1}{3}$
$\left.\begin{array}{l}
1/3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 &-13 & 43 & -13 \\\hline
& 1 & -4 & +13 \\\hline
3 & -12 & 39 & |\ \ 0 \end{array}$
$f(x)=(x+4)(x-\displaystyle \frac{1}{3})(3x^{2}-12x+39)$
$=(x+4)(x-\displaystyle \frac{1}{3})\cdot 3(x^{2}-4x+13)$
use the quadratic formula on $x^{2}-4x+13=0$
$x=\displaystyle \frac{4\pm\sqrt{16-4(13)}}{2}=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm 6i}{2}=2\pm 3i$
$f(x)=(x+4)(x-\displaystyle \frac{1}{3})(x-2+3i )(x-2-3i )$
Zeros: $-4,\displaystyle \ \frac{1}{3},\ 2-3i,\ 2+3i$
