n=3; -3 and 5 +5i are zeros ; f(2) =170

Question

find an nth-degree polynomial function with real coefficients satisfying the given conditions.

Answer 1

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.