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!
 A: 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 $-3-4i$.
A: In a first time try z=a where a is real number.
And so solve $2\,{z}^{3}+9\,{z}^{2}+17\,z+12=0$ and $6\,i{z}^{2}+3\,iz-9\,i=0$
You find a root.
A: HINT...you could start by speculatively assuming that one of the roots is real.
If such a root exists, then it must be a root of the cubic formed by the real part of your polynomial, and also a root of the quadratic formed by the imaginary part.
This reveals that, for example, $z=-\frac32$ is a root.
