Answer
$f(x)=(x-4)(x+7)(x+3i )(x-3i )$
Zeros: $\ \ 4,\ -7,\ 3i,\ -3i$
Work Step by Step
Degree $4$: there are $4$ complex zeros.
First, rational zero candidates: $\displaystyle \frac{p}{q}=$
$\displaystyle \frac{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 7,\pm 9,\pm 12,\pm 14,\pm 18,}{\pm 1}$ ,
$\displaystyle \frac{\pm 21,\pm 28,\pm 36\pm 42,\pm 63,\pm 84,\pm 126,\pm 252}{\pm 1}$
Trying synthetic division, ... $x-4$
$\left.\begin{array}{l}
4 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
1 & 3 &-19 & 27 & -252 \\\hline
& 4 & 28 & 36 & +252 \\\hline
1 & 7 & 9 & 63 & |\ \ 0 \end{array}$
$f(x)=(x-4)(x^{3}+7x^{2}+9x+63)$
... factor in pairs ...
$x^{3}+7x^{2}+9x+63=x^{2}(x+7)+9(x+7)=(x+7)(x^{2}+9)$
$x^{2}+9=0\Rightarrow x=\pm 3i$
$f(x)=(x-4)(x+7)(x+3i )(x-3i )$
Zeros: $4,\ -7,\ 3i,\ -3i$
