(Ax+ n) (x+m)
Am+n=b
c=n*m
So, for example, if you have 2(x2)+4x-32
(2x+8)(x-4)
A=2
m=-4
n=8
Am+n=(2*-4)+8=4
b=4
c=-4*8
c=32
That doesn't always work. Say you have 4x^2 + 16x + 15. You say to do (4x + something)(x + something) but that won't give you the answer. The answer is actually (2x + 3)(2x + 5).
The trick is to come up with all factors of A and C and unfortunately guess and test them. If you had 4x^2 + 16x + 15 you'd say the factors of A were +-1, 2, and 4 while the factors of C were +-1, 3, 5, and 15. Now you just need to guess and test. You can use some logic to solve it though...
Always consider the middle number because it can often help you to narrow down your choices. First of all I see that 15 is positive so I'm either working with 2 negatives or 2 positives for C. Since B is positive I know I must need 2 positives. That rules out all the negatives. So I'm going to have (?x + ?)(?x + ?).
If I use 15 and 1, they already add up to 16. But don't forget that A = 4 so I have to use factors of 4. They WON'T BOTH BE 1! Therefore if I use 15 and 1 to get my C, they will add up to be MORE than 16! So I want to use 3 and 5.
(?x + 3)(?x + 5)
From here I'm working with 1 and 4 or I'm working with 2 and 2. I see that 3 + 5 = 8. Double it and get 16. Since I can use 2 and 2 to get A, that must be correct. Alternatively I could have guess and tested. I would just have to test 1, 4; 4, 1; 2, 2.
The quadratic method can be awesome if you can do it quickly. It may even save you time when there are rational answers but really crappy high coefficients.
