there aren't enough terms to factor this by grouping; you need at least four, at least as far as i know? trinomials are usually pretty easily factored by trial and error
(2x - 3)(x - 1)
is how it factors down. you know that two only has two factors (2 and 1), so you always have (2x +_) and (1x +_) for trinomials beginning with 2x^2. from there, i usually just plug in factors of your constant (3, in this case) and see which ones fit the empty spaces!!
