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
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
3 answers
(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.
(x^2 + 16)(x^2-5)
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
something about "great minds ...." .lol
notice even the posing time was the same