Hi, I have an algebra question that I\'m having trouble with. I wrote it below.

Question

Hi, I have an algebra question that I'm having trouble with. I wrote it below.

"Write a 3rd-degree polynomial equation with real coefficients and the given roots. 2, 2, and i."

The answer key lists the answer as "x^3 - 6x^2 + 13x - 10 = 0"

This seems like an impossible answer to me as in order for the equation to have only real coefficients, (x - i) must exist with its complex conjugate (x + i). Also, 2 is a root of multiplicity 2 so I concluded that (x - 2)^2 exists as a factor(s) of the equation. This would mean my equation would have to be 4th degree (by the prime factorization theorem).

Can anyone tell me what I am missiing?

Thanks,

Kevin.

Explanation / Answer

roots. 2, 2, and i .Another conjugate pair root of i = -i

(x -2)(x -2)(x+i)(x -i) = (x^2 - 4)(x ^2 +1)

Your solution is correct , if theroots are 2 , 2 , i and -i

It is possible that x = 2, 2 is only one root x = 2

Check that

Otherwise its not possible to get 3rd degree with 4 roots

x^3 - 6x^2 + 13x - 10 = 0

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