Answer
$f(x)= (x-3i)(x+3)(x+4)$
Work Step by Step
If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x).
Synthetic division gives us a quotient q(x), with degree one less than f(x).
and we know the remainder is zero,
so $f(x)=(x-k)\cdot q(x).$
We use this first.
$k=3i$
$\left\{\begin{array}{lllllll}
3i & | & 1 & 7-3i & 12-21i & -36i & \\
& & & 3i & 21i & 36i & \\
& & -- & -- & -- & -- & --\\
& & 1 & 7 & 12 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x-3i)(x^{2}+7x+12)$
We now factor the trinomial $x^{2}+bx+c$ by searching for
two factors of $c$ whose sum is $b.$
Factors of $12$ whose sum is $7$ ... are $4$ and $3$:
$x^{2}+7x+12=(x+3)(x+4)$
So,
$f(x)= (x-3i)(x+3)(x+4)$