Why does this method for solving equations with complex number roots not always work?

Why does this method for solving equations with complex number roots not
always work?

Here's the question:
$1+3i$ is a root of the cubic $z^3+6z+20=0$. Identify the other two roots.
Obviously the conjugate complex number is another root so one is $1-3i$.
So I use the general formula $z^2-(\alpha+\beta)+\alpha\beta=0$, where
$\alpha$ and $\beta$ are the complex conjugate pair, to work out that the
cubic in question comes from, which I find to be:
$(z^2-z+10)(z+a)$, where $a$ is some number.
For the next part I set the cubic as $z^3+0z^2+6z+20=0$.
$\therefore$ $az^2-z^2=0z^2$
so $a=1$.
However, my problem arises when I try to find $a$ using the other powers:
With the same method,
$-za+10z=6z$
$\therefore a=4$
And with: $10a=20$, $a=2$.
Which value of $a$ is the correct one?
Surely I should get the same value for $a$ on each of these? Or am I
making a silly error?
For varying questions of this nature, I've found that some work fine and
some don't.