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?
(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.
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.
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