# Thread: Factorising a polynomail over c

1. ## Factorising a polynomail over c

How would one start this factorisation?

$P(z) = z^2 +(3 +2i)z +6i$

Would you just sub in values of z until you get P(z) to equal zero? Or is there a better way to approach this problem?

I subbed in values til P(z) equals zero and I found -3 to be a root, but is there a more sophisticated way to solving this problem than just subbing in values until you get a zero?

David.

2. In general, if $p(z)=az^2+bz+c\in\mathbb{C}[z]\quad (a\neq 0)$ then,

$p(z)=a(z-z_1)(z-z_2)$

where $z_1,z_2$ are the solutions of the quadratic equation $az^2+bz+c=0$ .

3. Oh yeah so it's
$z^2 + 3z + 2zi + 6i$
$z(z + 3) + 2i(z + 3)$
$(z +2i)(z + 3)$

thanks