Solve 2^2-5X+2=0
I forgot how to do this. I get x=4,1 but that's wrong. I thought you multiply 2*2 and then you know that you need a factor of 4 that adds up to 5 which are 4 and 1.
Then what?
We know that if this equation is factorable, that we will get the following form (Ax $\displaystyle \pm$ B)(Cx$\displaystyle \pm$D) = 0
He first took $\displaystyle 2x^2$ and factored it. 2 is a prime number so, it's factor's are just 2 and 1, so we know what A and C should be.
(2x $\displaystyle \pm$ B)(1x$\displaystyle \pm$D) = 0
Using FOIL, we know that the last two terms will be multiplied together to come up with a constant. In our case, the constant is 2. Again it is a prime. So B and D are going to be a combination of 1 and 2. Since the last constant in the original equation is a positive number, we know that when multiplied together using FOIL B and D will be both positive or both negative.
Hope that helps some
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Solving Quadratic Equations: Solving by Factoring