We already learned this in class but I still don't understand.
It's basically the opposite of solving a problem that looks like this (4x+5)(3x-2) and getting 12x+15x-8x-10 which you reduce to a polynomial which would be 12x^2+7x-10. You start with the polynomial, and have to find the factors which will be in the form I started with. So, if your problem is 12x^2+7x-10, I see it as having a middle number, a front number and an end number. You need to find numbers that can be multiplied to get 12, the front number, and numbers that multiply to get -10, the end number. It is important to keep track of you negatives--the best thing is to think of something like 7x-10 as 7x + (-10) so you don't forget it is negative! Now, 3x4 and 2x6 and 1x12 are all 12, and 2x5 and 1x10 are both 10, but you need to pick the ones that when multiplied and added will equal your middle number. Try putting them in the answer format. (2x+1)(6x-10) will give you 12x^2-20x+6x-10 or 12x^2-14x-10, which is not what you started with
(x-5)(12x+2) will give you 12x^2+12x-60x-10, or 12x^2-58x-10, which is also wrong
so you will end up going with(4x+5)(3x-2) eventually.