1. ## 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!

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

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

$\begin{array}{r|rrrr} & {1} & {1-i4} & {17-i16} & {-87-i116}\\ \, \\{3+i4} & & {3+i4} & {12+i16} & {87+i16}\\\hline & 1 & 4 & 29 & 0\end{array}$

3. ## Re: Cubic polynomial - complex coefficients HELP ME PLEASE

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 $3+4i$ to reduce it to a quadratic that you can solve.