Best Method for Factoring this Expression I'm wondering if anyone has an opinion on the best way of factoring the following expression. As one can see it is quite complicated.

$$(15x^2)(x^3+{4})^{4}(1-2x^{2})^{3}+(12x)(x^{3}+4)^{5}(1-2x^{2})^{2}\over (1-2x^{2})^{6}$$

Any help is of course extremely appreciated
There isn't really a best way, but here's how I would do it (sorry in advance for the messy work, I'm bad at formatting my answers :/).

First, analyze the entire expression and look for like terms.
$(x^3+4)$,  $x$, and  $(1-2x^2)$  are like terms.
Now, try to factor these like terms out (to the highest possible exponent).

$$ \frac{(x^3+4)^4⋅(1−2x^2)^2⋅x⋅[5⋅3x(1−2x^2)−(x^3+4)⋅3⋅(−4)]}{(1−2x^2)^6}$$

Notice that the denominator, $(1−2x^2)^6$, has the same term as the top, $(1−2x^2)^2$
Thus, you can divide the top term by the bottom term, cancelling the top term to get:

$$ =\frac{(x^3+4)^4⋅x⋅[5⋅3(1−2x^2)−(x^3+4)⋅3⋅(−4)]}{(1−2x^2)^4}$$

Now, simplifying some of the coefficients to get:

$$ =\frac{(x^3+4)^4⋅x⋅[15x(1−2x^2)−(x^3+4)⋅(-12)]}{(1−2x^2)^4}$$

You can expand the terms inside of the brackets to combine the terms inside and further simplify the expression.

$$=\frac{(x^3+4)^4⋅x⋅[15x−30x^3+12x^3+48]}{(1−2x^2)^4}\\
 =\frac{(x^3+4)^4⋅x⋅[−18x^3+15x+48]}{(1−2x^2)^4}$$

Notice that the terms inside the brackets all share a common factor: 3. Factor 3 out and you get:

$$=\frac{3x(x^3+4)^4(−6x^3+5x+16)}{(1−2x^2)^4}$$

And this is your final answer. You can possibly factor out a negative from $(−6x^3+5x+16)$ if you wanted to, but it's not necessary.
I'm not really sure if you're more interested in the final result, or on how to achieve it. Anyway, Mathematica factors it as
\begin{equation}
-\frac{3x(4+x^3)^4(-16-5x+6x^3)}{(-1+2x^2)^4}.
\end{equation}
Apologies if you were after something with more detail (and you could of course verify this result yourself using WolframAlpha).