# Find roots of cubic polynomial with complex coefficient

## Homework Statement

Find roots of
$$-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0$$

## The Attempt at a Solution

I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of ##\lambda + z## between them,
$$-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i$$
It doesn't seem to work.
$$-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Find roots of
$$-\lambda ^3 +(2+2i)\lambda^2-3i\lambda-(1-i) = 0$$

## The Attempt at a Solution

I tried my old trick
I tried to separating the 4 terms into 2 pairs and try to find a common factor in the form of ##\lambda + z## between them,
$$-\lambda ^2 (\lambda -2-2i) - 3i\lambda -1+i$$
It doesn't seem to work.
$$-\lambda (\lambda^2 +3i) - (2+2i)\lambda -1+i$$
Nope.
And I am out of tricks.
Should I keep trying to factorize it or there is something else I should do about polynomial with complex coefficient?

You can use Cardano's formula for the solution of a cubic. It works for any coefficients, not just for real ones.

BearY
LCKurtz