Question

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

Answer 1

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