Answer
$f(x)= [x-(1+i)](x+3)(2x-1)$
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.
$k=1+i$
$\left\{\begin{array}{lllllll}
1+i & | & 2 & 3-2i & -8-5i & 3+3i & \\
& & & 2+2i & 5+5i & 3+3i & \\
& & -- & -- & -- & -- & --\\
& & 2 & 5 & -3 & \fbox{$0$} &
\end{array}\right.$
$f(x)= [x-(1+i)](2x^{2}+5x-3)$
We now factor the trinomial $ax^{2}+bx+c$ by searching for
two factors of $ac$ whose sum is $b.$
If they exist, rewrite $bx$ and factor in pairs.
Factors of $2(-3)=-6$ whose sum is $+5$ ... are $+6$ and $-1$:
$2x^{2}+5x-3\\=2x^{2}+6x-x-3$
$=2x(x+3)-(x+3)$
$=(x+3)(2x-1)$
So,
$f(x)= [x-(1+i)](x+3)(2x-1)$
