Absolute convergence of series variant of the geometric series

So I want to prove whether the following series converges absolutely or not:

$$\frac{1}{2}\sum_{n=0}^\infty (n^2+3n+2)q^n$$

where $$q \in \mathbb{C}, \mid q\mid<1.$$

My attempt was:

$$\frac{1}{2}\sum_{n=0}^\infty \mid(n^2+3n+2)q^n\mid = \frac{1}{2}\sum_{n=0}^\infty \mid(n^2+3n+2)\mid \mid q^n\mid$$

It holds that $$n \geq 0 \Rightarrow n^2+3n+2 > 0 \,\, \forall n$$ therefore,

$$\Rightarrow \frac{1}{2}\sum_{n=0}^\infty \mid(n^2+3n+2)\mid \mid q^n\mid = \sum_{n=0}^\infty (n^2+3n+2)\mid q^n\mid$$

For showing that $$\frac{1}{2}\sum_{n=0}^\infty (n^2+3n+2)\mid q^n\mid$$ converges it is sufficient to show that $$\sum_{n=0}^\infty (n^2+3n+2)\mid q^n\mid$$ converges.

To be continued...

My problem is now that I don't really know which convergence criterion I should use nor I know what to do with that $$\mid q^n \mid$$ part whether I can omit the absolute value or not (my guess is not since it is a complex number and $$\mid .\mid \;: \mathbb{C} \rightarrow \mathbb{R}$$ and that basically everything.