# How do you factor  3n^2+2n-1?

$3 {n}^{2} + 2 n - 1 = \left(3 n - 1\right) \left(n + 1\right)$
I found this by recognising that if the original quadratic has factors with integer coefficients then they must be of the form $\left(3 n \pm 1\right)$ or $\left(n \pm 1\right)$ in order that the leading and trailing coefficients be $3$ and $- 1$.