SOLUTION: Degree 5; zeros:-2;-i;3+i

If one zero is A+Bi, then A-Bi is also a zero.

So since one zero is -i, then +i is also a zero.

And since one zero is 3+i, then 3-i is also a zero.

So these are the 5 zeros:

   x=-2,  x=-i;   x=+i;   x=3+i;   x=3-i

Get 0 on the right of all 5 equations:

 x+2=0;  x+i=0;  x-i=0; x-3-i=0; x-3+i=0

Multiply both sides of all those 5 equations together:

 (x+2)(x+i)(x-i)(x-3-i)(x-3+i) = 0

Multiply all those together (it's a lot of work) and set f(x)
equal to what you get.  If you have trouble, tell me in the 
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Edwin