The remaining zero is the conjugate of 5+5i=5-5i.
(x-5-5i)(x-5+5i)=(x-5)2+25=x2-10x+50.
The polynomial (cubic) is (x+3)(x2-10x+50)=
x3-10x2+50x+
3x2-30x+150=x3-7x2+20x+150 or multiple n thereof.
If f(2)=170, 23-7×22+20×2+150=8-28+40+150=198-28=170, therefore:
170n=170, n=170/170=1
Polynomial is x3-7x2+20x+150.