How do you factor #2n^2+n-6#?

1 Answer
Jul 21, 2016

Answer:

(2n - 3)(n + 2)

Explanation:

Use the new AC method (Socratic Search)
#y = 2n^2 + n - 6 = 2(n + p)(n + q) Converted trinomial #y' = n^2 + n - 12#
Find 2 numbers knowing sum (1) and product (- 12). They make the factor pair (-3, 4).
Back to trinomial y, p = -3/2 and q = 4/2 = 2
Factored form of y:
y = 2(n - 3/2)(n + 2) = (2n - 3)(n + 2)