Find the roots of $P( z) =2z^{3} +( 9+6i) z^{2} +( 17+3i) z+12-9i$

I need to get all the roots of $P( z) =2z^{3} +( 9+6i) z^{2} +( 17+3i) z+12-9i$ but I dont know how to get the first factor!

Guessing numbers I got that $i$ is one of the roots but is there any systematic way to get the roots? Thanks!

Here's what I would do. Assume that the polynomial has at least one real root (this might not necessarily be true). Then we have

$$P(z)=2z^{3} +( 9+6i) z^{2} +( 17+3i) z+12-9i$$

$$=(2z^3+9z^2+17z+12)+i(6z^2+3z-9)=R(z)+i I(z)$$

Solving the polynomial $I(z)$ gives us the roots

$$I(z)=0\Rightarrow z=-\frac{3}{2}\text{ or }z=1$$

Testing both of these with the real polynomial gives

$$R(1)=40\text{ and }R(-3/2)=0$$

Alright, so we have one root. We can then assume the polynomial is of the form

$$P(z)=\left(z+\frac{3}{2}\right)(az^2+bz+c)$$

Expanding and solving gives us

$$a=2$$

$$b=6+6i$$

$$c=8-6i$$

This gives us a quadratic

$$f(z)=2z^2+(6+6i)z+(8 - 6 I)$$

Solving this using the quadratic formula gives us the other roots $i$ and $-3-4i$.