Adding fractions with variables and using common denominator. Merging and shortening $\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}$

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So I have these fractions I need to merge together and shorten:

$$\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}$$

I understand that I need to make the denominators the same so that I can merge these fractions together, but I have no idea how.

I assume I need to find the common denominator, which is a friend told me was $2a^2 - 64$. I don't know what to do from here, please help?

How do I use the common denominator to make these denominators the same?

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Consider that this expression can be rewritten as:

$$\frac{1}{2a+8} + \frac{4}{a^2-16} + \frac{4}{a-4}=\frac{1}{2(a+4)} + \frac{4}{a^2-16} + \frac{4}{a-4}$$

Now $(a+4)(a-4)=a^2-16$, so multiply fractions by clever forms of one:

$$\frac{1}{2(a+4)}\frac{(a-4)}{(a-4)} + \frac{4}{a^2-16}\frac{2}{2} + \frac{4}{a-4}\frac{2(a+4)}{2(a+4)}=\frac{(a-4)}{2(a^2-16)} + \frac{8}{2(a^2-16)} + \frac{8(a+4)}{2(a^2-16)}$$

Now, we can simply add up fractions:

\begin{align}\frac{(a-4)}{2(a^2-16)} + \frac{8}{2(a^2-16)} + \frac{8(a+4)}{2(a^2-16)}&=\frac{8a+32+8+a-4}{2(a^2-16)}\\\\ &=\frac{9a+36}{2(a^2-16)}\\\\ &=\frac{9(a+4)}{2(a+4)(a-4)}\\\\ &=\frac{9}{2(a-4)}\end{align}