Asked by Lynn
I am trying to find the real coefficients of the given roots. Did I do this right?
3,-9,3+2i
(x-3)(x+9)[x-(3+2i)][x-(3-2i)]=0
(x-3)(x+0)[(x-3)+2i][(x-3)-2i]=0
(x-3)(x+9)[(x-3)^2-2i^2]=0
(x-3)(x+9)(x^2-6x-8)=0
(x^4-71x^2+114x+216)=0
answer x^4-71x^2+114x+216=0
3,-9,3+2i
(x-3)(x+9)[x-(3+2i)][x-(3-2i)]=0
(x-3)(x+0)[(x-3)+2i][(x-3)-2i]=0
(x-3)(x+9)[(x-3)^2-2i^2]=0
(x-3)(x+9)(x^2-6x-8)=0
(x^4-71x^2+114x+216)=0
answer x^4-71x^2+114x+216=0
Answers
Answered by
Anonymous
Hmm.. I never done this before but I think you are right for the most part. You made a mistake on (x-3)(x+9)(x^2-6x-8) = 0. (x-3)^2 - 2i^2 does not expand to x^2-6x-8.
x-3)^2 - 2i^2
x^2 - 6x + 9 + 2
x^2 - 6x + 11
x-3)^2 - 2i^2
x^2 - 6x + 9 + 2
x^2 - 6x + 11