How do you factor $3 n^{2} + 2 n - 1$ $3n^2+2n-1$ ?

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How do you factor $3 n^{2} + 2 n - 1$ $3n^2+2n-1$ ?

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Answer 1 · 2015-06-05T19:36:19

$3 n^{2} + 2 n - 1 = (3 n - 1) (n + 1)$ $3n^2+2n-1 = (3n-1)(n+1)$

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

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How do you factor $3 n^{2} + 2 n - 1$ $3n^2+2n-1$ ?

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