When factoring, we look for something called a common factor; in other words, something that appears in all of the terms. Right off the bat, we can see that all of our coefficients (the numbers next to the #x#s) are even - so 2 is a common factor. In addition, all of our terms contain an #x#. Putting it all together, our first common factor is #2x#:

#2x(x^2+5x+6)#

If we were to distribute the #2x#, we would get #2x^3+10x^2+12x#, which is our original problem.

Now, we proceed to factoring the quadratic equation #x^2+5x+6#. We do this by looking for two numbers that add to 5 and multiply to 6; these numbers are 3 and 2. If you're wondering why: when we factor a quadratic like this one, we want it in the form #(x+a)(x+b)#, where #a# and #b# multiply to the constant term and add to the first-degree term (in this case, our constant term is #6# and our first degree term is #5x#). Here, we have #a = 3# and #b = 2# (#a = 2# and #b = 3# works too). So, the factored form of the quadratic equation #x^2+5x+6 = (x+3)(x+2)#. Back to the main problem.

Now that we have our quadratic completely factored, we can finish up. We simply replace our quadratic with the factored version, like this:

#2x(x+3)(x+2)#

Since we can't simplify this any further, we can say that #y = 2x^3+10x^2+12x# in factored form is #y = 2x(x+3)(x+2)#.