Form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros.

The zeros of the equation are 8+i and 7. however since this is a third degree polynomial, the other root missing is the conjugate of the imaginary term 8-i, which is 8 + i. We multiply then (x - 8 - i) * (x - 7) * (x - 8 + i). The expansion is (x2 - 16x + 1 - - 1) * (x-7) = (x2 - 16x + 2) (x-7) = x3 - 7 x2 - 16x2 + 112 x + 2x - 14 = x3 - 23 x2 + 114 x - 14