Dividing polynomial fractions with varying term quantities I'm working through an old algebra book as a refresher and I've come across what should be a simple polynomial division. The exercise prompts the reader to perform the following operation:
$$
\frac{a^2-9}{a^2+3a} \div \frac{a-3}{4}
$$
I started off by inverting the $\div$ sign by instead multiplying by the reciprocal which results in:
$$
\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}
$$
I spent nearly 2 hours at this point trying everything my mind could conjure in terms of factoring, simplifying, and multiplying, but none of my attempts ever arrived at the listed answer:
$$
\frac{4}{a}
$$
If someone could help me through the steps required to solve this, you'll have taught a man to fish.
Solution from below:
$$
\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}=\frac{4(a-3)(a+3)}{a(a+3)(a-3)}=\frac{4}{a}
$$
I had not factored $a^2 + 3a$ to $a(a+3)$ correctly in any of my attempts.

Because $$\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}=\frac{4(a-3)(a+3)}{a(a+3)(a-3)}=\frac{4}{a}
$$
