Question 745584
(x-3), (x-2i) and (x+2i) tells us...

{{{p=a(x-3)(x-2i)(x+2i)}}} 

for some constant a


{{{(x-2i)(x+2i)}}}

...is a difference of squares - with a twist:


{{{(x-2i)(x+2i)=x^2+4}}}


so we now use distribution...

{{{p=a(x-3)(x^2+4)=a(x(x^2+4)-3(x^2+4))}}}


{{{p=a(x^3+4x-3x^2-12))}}}



{{{p=a(x^3-3x^2+4x-12))}}}

with the principal polynomial

{{{x^3-3x^2+4x-12}}}