# Cubic polynomial - complex coefficients HELP ME PLEASE

• Jan 10th 2013, 03:23 PM
dynamic
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!
• Jan 10th 2013, 03:35 PM
russo
Re: Cubic polynomial - complex coefficients HELP ME PLEASE
You just apply Ruffini with the root you have. You end up with $\displaystyle P(z) = z^2 + 4z + 29$

$\displaystyle \begin{array}{r|rrrr} & {1} & {1-i4} & {17-i16} & {-87-i116}\\ \, \\{3+i4} & & {3+i4} & {12+i16} & {87+i16}\\\hline & 1 & 4 & 29 & 0\end{array}$
• Jan 10th 2013, 03:40 PM
Plato
Re: Cubic polynomial - complex coefficients HELP ME PLEASE
Quote:

Originally Posted by dynamic
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 $\displaystyle 3+4i$ to reduce it to a quadratic that you can solve.