Question 380012
Suppose our polynomial is P(x). If P(x) has roots 2i and 8i, then P(x) = (x-2i)(x-8i)Q(x), where Q(x) is an arbitrary polynomial. If Q(x) = 1 (to minimize the degree), then {{{P(x) = (x-2i)(x-8i) = x^2 - 10ix - 16}}}. The only issue is that we have a non-real coefficient, 10i.

Suppose we wanted to find a polynomial with only real coefficients.

I'll use a trial and error case: By Vieta's formulas, we want the sum and the product of the roots to be real numbers. Therefore I'll introduce the conjugates of the first two roots, -2i and -8i. Our polynomial becomes

(x-2i)(x+2i)(x-8i)(x+8i)

= {{{(x^2 + 4)(x^2 + 64)}}}
= {{{x^4 + 68x^2 + 256}}}

Either answer could be correct, depending on whether the coefficients are assumed to be real.