Question

Can someone tell me how they got the answer?

Find a polynomial equation with real coefficients that has the given roots.
4i, sqrt5
my answer: x^4-22x^2+80=0
correct answer: x^4+11x^2-80=0

Answer 1

If 4i is a root, there must have been a -4i
and if √5 was a root there must have been a -√5

so the factors were (x-4i)(x+4i)(x+√5)x-√5)
= (x^2 + 16)(x^2-5)
= x^4 - 5x^2 + 16x^2 - 80
= x^4 + 11x^2 - 80

Answer 2

(x+4i)(x-4i)(x-sqrt5)(x+sqrt5)
(x^2 + 16)(x^2-5)
x^4 +11x^2 - 80


I don't know what you did.

Answer 3

What is that saying?
something about "great minds ...." .lol

notice even the posing time was the same