Prove that for $x\in\Bbb{R}$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$. The problem I have is:

Prove that for real numbers $x$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.

Since there aren't really any similar examples in my book, I've been unsure how to first approach this problem and have been trying to use to find similar questions.

Do I try and divide the problem into regions, like at absolute value inequalities, using $x^2-2x-15=(x-5)(x+3)$

Or could I just simply have two cases, those being $x\ge 3$ and $x\lt3$ like at How to write an expression in an equivalent form without absolute values?
We know that $|x|&lt;3$. Therefore $-3&lt;x&lt;3$, subtracting $5$ from each side gives:

$-8&lt;x-5&lt;-2$

And so:

$|x-5|&lt;8$

This implies that:

$|x^2-2x-15|=|(x-5)(x+3)|=|x-5|\cdot |x+3|&lt;8|x+3|$

As needed.
Implies means inclusion of sets of solutions. 

Therefore, we could solve both inequalitites and show their sets of solutions are one inside the other.

For $|x|&lt;3$, it is $(-3,3)$.

For $|x^2-2x-15|&lt;8|x+3|$, we get $|(x-5)(x+3)|=|(x-5)(x+3)|&lt;8|x+3|$. This is the same as $x\neq-3$ and $|x-5|&lt;8$. This is $[5-8,5+8]=(-3,13]$.

Now we can simply check that $(-3,3)\subset(-3,13)$