Algebra
posted by Joey .
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^422x^2+80=0
correct answer: x^4+11x^280=0

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 (x4i)(x+4i)(x+√5)x√5)
= (x^2 + 16)(x^25)
= x^4  5x^2 + 16x^2  80
= x^4 + 11x^2  80 
(x+4i)(x4i)(xsqrt5)(x+sqrt5)
(x^2 + 16)(x^25)
x^4 +11x^2  80
I don't know what you did. 
What is that saying?
something about "great minds ...." .lol
notice even the posing time was the same
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