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

Question

Julia Adams · Accepted Answer

#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 #(3n+-1)# or #(n+-1)# in order that the leading and trailing coefficients be #3# and #-1#.

Avery Alexander · Answer

undefined To factor the quadratic expression $3n^2 + 2n - 1$, you can use the quadratic formula or factorization by grouping. Using the latter method:

1. Multiply the leading coefficient (3) by the constant term (-1) to get -3.
2. Find two numbers that multiply to -3 and add up to the coefficient of the linear term (2). These numbers are 3 and -1.
3. Rewrite the quadratic expression using these numbers: $3n^2 + 3n - n - 1$.
4. Group the terms: $(3n^2 + 3n) + (-n - 1)$.
5. Factor out the greatest common factor from each group: $3n(n + 1) - 1(n + 1)$.
6. Factor out the common binomial factor $(n + 1)$: $(3n - 1)(n + 1)$.

So, the factored form of $3n^2 + 2n - 1$ is $(3n - 1)(n + 1)$.