undistribute the x and 4 you have a simpler quadratic polynomial.

4x[6x^2 + 7x - 20]

That further breaks down into (2x+5)(3x-4) using the reverse foil method to end with:

4x[(2x+5)(3x-4)]


http://www.jiskha.com/mathematics/algebra/foil.html

Need more?

well, you can factor out 4x from each part

(4x)(6x^2 + 7x - 20)

then um... I used to be good at doing integer factoring, but you could probably use the quadratic equation if you're stumped.

Equals... what?

I tried it out as well, easily getting to where Yippee did, and then it took a bit of finnessing to arrive at where xavier did.

In order to help you understand it, first you want to factor out all the major common factors - i.e. if the entire thing is divisible by 2 (or in this case, 4), and if there's a common factor of x in there.

Next, looking at the factored down quadratic, if the x^2 term does not have an integer in it, then it is easy to figure what combination of things multiply to give you the constant term and add (or subtract) to give you the single-power term.

Unfortunately, we've got an integer for the x^2 term, so we have that many more sets of possibilities, even more so because 6 isn't a prime number, so it could either be (x+/-#)(6x+/-#) or (2x+/-#)(3x+/-#). Keeping a few other shortcuts in mind:
- the last term is negative, so one factor is addition, the other factor is subtraction
- an even number times anything is even, and only an odd number times another odd number is odd.
- the middle term is odd, and since only the difference between even and odd numbers is odd, then that gives a hint as to what the multiplied single-powers will break down into (i.e. combine this with the multiplication trick).