Question 1157875
2 and 4i and -4i are all zeros, since complex roots always occur in conjugate pairs.
f(1)=-17
f(x)=0 ? or f(x)=.  
nothing can mean a blank, or it can mean a zero.

In any case, the factors are (x-2)(x-4i)and (x+4i)
(x^2-16i^2)=(x^2+16) is another factor, when I multiply the second and the third together.

So the polynomial is (x-2)(x^2+16)=x^3-2x^2+16x-32
put in the form a(x^3-2x^2+16x-32), since it may be a multiple of that.
f(1)=-17
so a(1-2+16-32)=-17
a(-17)=-17, so a=1
the answer is x^3-2x^2+16x-32

{{{graph(300,300,-20,20,-100,125,x^3-2x^2+16x-32)}}}