Answer
$f(x)= (x+2-i)(2x+1)(3x+2)$
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.
$k=-2+i$
$\left\{\begin{array}{lllllll}
-2+i & | & 6 & 19-6i & 16-7i & 4-2i & \\
& & & -12+6i & -14+7i & -4+2i & \\
& & -- & -- & -- & -- & --\\
& & 6 & 7 & 2 & \fbox{$0$} &
\end{array}\right.$
$f(x)= [x-(-2+i)](6x^{2}+7x+2)$
$=(x+2-i)(6x^{2}+7x+2)$
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 $6(2)=12$ whose sum is $+7$ ... are $+4$ and $+3$:
$6x^{2}+7x+2\\=6x^{2}+3x+4x+2$
$=3x(2x+1)+2(2x+1)$
$=(2x+1)(3x+2)$
So,
$f(x)= (x+2-i)(2x+1)(3x+2)$
