Find a polynomial function with real coefficients that has the given zeros. (there are many correct answers.) 8, 8, 1 + i

If 1+i is a root then 1-i must be a root because complex roots come in conjugate pairs, so
(x-8)² * (x-1-i) * (x-1+i) = 0
(x-8)² * (x²-x+xi-x+1-i-xi+i+1)=0
(x-8)² * (x²-2x+2) = 0
(x²-16x+64)(x²-2x+2)=0
x⁴-2x³+2x²-16x³+32x²-32x+64x²-128x+128=0
f(x)=x⁴-18x³+98x²-160x+128

if I didn't lose track... that was tough on a phone with me paper to scratch on!

Bc equals mb that should be your answer