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Question 997738: I need to see the steps to solve:
Find the cubic equation that has -1 and 2i as roots.
Thank you.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! imaginary or complex roots come in pairs as best i can remember.
your roots are x = -1, x = +2i, x = -2i
your factors are (x+1), (x-2i), (x+2i)
(x - 2i) * (x + 2i) = x^2 + 2ix - 2ix -4i^2
the + 2ix and the - 2ix cancel out and you are left with:
x^2 - 4i^2
since i^2 = -1, you are left with:
x^2 -4*-1 which is equal to:
x^2 + 4
now you multiply this by (x + 1) to get:
(x + 1) * (x^2 + 4) = x^3 + 4x + x^2 + 4
combine like terms and order the terms in descending order to get:
x^2 + x^2 + 4x + 4
that should be your cubic equation.
here's a reference on complex roots.
there are numerous others on the web.
https://www.mathsisfun.com/algebra/fundamental-theorem-algebra.html
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