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Math Help - Cubic polynomial - complex coefficients HELP ME PLEASE

  1. #1
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    Thumbs down Cubic polynomial - complex coefficients HELP ME PLEASE

    Hello people!

    Help would be greatly appreciated... z3 + (1 - 4i)z2 + (17 - 16i)z + (-87 - 116i) = 0, where z = 3 + 4i is one of the roots...

    step by step working would be fandabadosy too; I know it's pretty basic but I've gone COMPLETELY blank

    Cheers!
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  2. #2
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    Re: Cubic polynomial - complex coefficients HELP ME PLEASE

    You just apply Ruffini with the root you have. You end up with P(z) = z^2 + 4z + 29

    \begin{array}{r|rrrr} & {1} & {1-i4} & {17-i16} & {-87-i116}\\ \, \\{3+i4} &  & {3+i4} & {12+i16} & {87+i16}\\\hline  & 1 & 4 & 29 & 0\end{array}
    Last edited by russo; January 10th 2013 at 03:40 PM.
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    Re: Cubic polynomial - complex coefficients HELP ME PLEASE

    Quote Originally Posted by dynamic View Post
    Hello people!

    Help would be greatly appreciated... z3 + (1 - 4i)z2 + (17 - 16i)z + (-87 - 116i) = 0, where z = 3 + 4i is one of the roots...

    step by step working would be fandabadosy too; I know it's pretty basic but I've gone COMPLETELY blank

    This is not a homework service

    You can find the exact answer on the web.

    To do it for yourself, divide the polynomial by 3+4i to reduce it to a quadratic that you can solve.
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