A fourth-degree polynomial can have -5,-8,4i and 5 as its zeros? why?

Answer 1

A 4th degree polynomial with zeros -5, -8, 4i, and 5 is equal to the following polynomial.

f(x) = (x + 5)(x + 8)(x - 4i)(x - 5)
f(x) = (x^2 + 13x + 40)(x^2 - 5x - 4ix + 20i)

So long as we don't have all real coefficients, we can have this 4th degree polynomial. Otherwise, we cannot, because the complex zero, 4i, does not have its conjugate among the other zeros.

Answer 2

Out of the four roots given, only one is unreal, and it is not accompanied by its conjugate (conjugate of a+ib is a-ib).

In general, polynomials have only real coefficients. In such polynomials, complex roots ALWAYS occur in conjugate pairs. As this is not the case here, it is not possible.

I think this is the answer ypu're looking for. Hope it helps!!