Question about factorization of quadratic expression and roots

It depends on what you mean by 'not being able to find' those two numbers! Some quadratics include numbers that are incredibly tough to factor - for instance, x² + 2x - 399 = 0. I wouldn't want to be in the position of finding two numbers that multiply to -399, and add to 2. Nonetheless, those numbers exist - that quadratic factors to (x + 21)(x - 19) = 0. It would be easy to see a problem like that, and assume, after trying for a while, that those numbers didn't exist. Be careful about making that assumption.

If you see a quadratic that you're struggling to use this technique on, there's probably something else you could do. Try the quadratic formula. Or, try going back in the problem and seeing if you could simplify it differently. In the example I just gave, this is a much easier solution (the example, by the way, is actually adapted slightly from an Official Guide problem):

x² + 2x - 399 = 0
x² + 2x + 1 = 400
(x + 1)(x + 1) = 400
x + 1 = +/- 20
x = -21 or +19

chetan2u is right, too. There are some quadratics for which a solution just doesn't exist. And some quadratics only have one solution. You'll see the latter on the GMAT sometimes, but I'm not sure that I've ever seen the former.

Here's an example of a quadratic with no real solutions:

x² + x + 4 = 0

And here's an example of a quadratic with one solution:

x² + 4x + 4 = 0