Determine 4 solutions of a complex polynomial

Hello fairy math helpers
I am trying to figure out the best way to approach this question.

1. Determine all solutions to $z^4 +2z^3 +6z^2 +8z +8 = 0$ given that one of the solutions is purely imaginary.

So I thought about factorising into two quadratics and then $4$ linear equations. From there I can figure out how to find the solutions using ($az +b$) I think? The problem is that I don't know how to factorise something with $4$ degrees before and our lecturer didn't give us an example.

Please help.

Answer

If there is a pair of roots such that is purely imaginary then $(z^2 + a^2)$ divides the polynomial.

Collect the even terms

$z^4 + 6 z^2 + 8$

and the odd terms

$2z^3 + 8z$

If $(z^2+a^2)$ divides your polynomials, a must be a root of both polynomials.

$a^2 = 4$

$(z+2i)(z-2i)$ are roots.

update:

What do I mean about dividing though by $(z^2 + a^2)$?

If $(z^2+ a^2)$ is a factor of $p(x)$ then

$p(x) = (z^2 + a^2) \frac {p(x)}{z^2 + a^2} = (z^2+a^2) q(x)$

$(z^2 + a^2)$ divides $p(x)$ leaving no remainder.

On splitting the even and odd terms:

$q(x)$ can be broken into an even polynomial and an odd polynomial

$q(x) = e(x) + o(x)$

$p(x) = (z^2 + a^2) (e(x) + o(x)) = (z^2 + a^2)e(x) + (z^2 + a^2) o(x)$
and

$(z^2 + a^2)e(x)$ is an even polynomial

$(z^2 + a^2)o(x)$is an odd polynomial

$(z^2 + a^2)$ is a factor of the strictly even terms of $p(x)$
and it is a factor of the strictly odd terms of $p(x)$