+9665 I always factor trinomials that are not perfect squares in a different way as yours, I need steps though, I do it in 1 step.
\(2x^2 + 11x + 5\\ = (ax + b)(cx + d) \text{Where } ac = 2,\;bc + ad = 11,\;bd = 5\\ = (2x + 1)(x+5)\)
The (ax + b)(cx + d) step is my thinking method, I usually write 2x^2 + 11x + 5 = (2x + 1)(x + 5).
