[SubAtomic]

(Original post by aznkid66)
Hm? Method? I think you misunderstood, it wasn't an alternative to factorization or the quadratic formula.

I just make all the coefficients integers before I factorize, or "have a stab." You should find that quadratics with rational roots are pretty easy to factorize as long as the leading coefficient and constant term are small integers.

So the answer you gave for 2 ends up being not so nice because 10 is not really that small for a leading coefficient when factorising such a quadratic?

[aznkid66]

Well, I know that the prime factorization of 10 is 2*5, and the constant term is also prime (2), so there are only 7 combinations to check. Better 10 than, say, 8.