Answer
$Q(x)=x^{3}+3x^{2}+4x-12$
Work Step by Step
The degree is $3$, there are $3$ complex zeros,
$\pm 2i$ are conjugates, (both are zeros)
$(x-3), (x-2i),\ (x+2i$) are all factors.
So,
$Q(x)=a(x-3)(x-2i)(x+2i)$
... taking a=1, apply difference of squares...
$Q(x)=(x-3)(x^{2}-4(-1))=(x-3)(x^{2}+4)$
... FOIL ...
$Q(x)= x^{3}+4x-3x^{2}-12$
$Q(x)=x^{3}+3x^{2}+4x-12$
