$$(3x^2 + 6x) / (x^3 - 4x)$$
To find the hole(s) of a rational expression, I must first create a common factor in both the numerator and denominator; however, can't come up with anything for the above expression.
I should add that I am aware of the common factor of $x$, but was told that there is a second common factor; thus, a second hole.
On
Factoring polynomials is analogous to factoring integers. For example, when you see $12-4$, this can be written as $4(3-1)$ since $4\cdot 3 = 12$ and $4\cdot -1 = -4$, and you recover the original expression $12-4$ (this is a good technique to check you answer, once you factor check by distributing to see that you recover the original expression). Now what allowed us to "factor-out" the $4$? It was because it was present in the factorization of both summands, the $12$ and the $-4$. Similarly, with an expression like $x^2+x$, we look for common things in each summand, i.e. an $x$ (treating the variable $x$ as we did with the integer $4$) and we have $x^2+x = x(x+1)$. Try this with your numerator and denominator.
$3x^2 + 6x = x(3x+6) = 3x(x+2)$ as I think you've noticed. What of the denominator? Well the denominator is $x^3-4x = x(x^2-4)$. This is called a difference of two squares. In general, $a^2-b^2 = (a+b)(a-b)$ which you can check by multiplying $(a+b)$ and $(a-b)$. Can you see where to go from here?