Answer
$f(x)= (x+i)(x+1)(2x+1)$
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.
$k=-i$
$\left\{\begin{array}{rrrrrrr}
-i & | & 2 & 3+2i & 1+3i & i & \\
& & & -2i & -3i & -i & \\
& & -- & -- & -- & -- & --\\
& & 2 & 3 & 1 & \fbox{$0$} &
\end{array}\right.$
$f(x)= (x+i)(2x^{2}+3x+1) $
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 $2(1)=1$ whose sum is $+3$ ... are $+2$ and $+1$:
$2x^{2}+3x+1\\=2x^{2}+2x+x+1$
$=2x(x+1)+(x+1)\\=(x+1)(2x+1)$
So,
$f(x)= (x+i)(x+1)(2x+1)$
