How do you factor #2x^2+ 3x - 14#?

1 Answer
Jun 28, 2015

Answer:

Factor #y = 2x^2 + 3x - 14#

Explanation:

f(x) = 2(x - p)(x = q)
I use the new AC Method (Google, Yahoo Search)
Converted #f'(x) = x^2 + 3x - 28 = (x - p')(x - q').#
Factor pairs of (-24) -> (-2, 12)(-4, 7). This sum is 3 = b
p' = -4 and q' = 7
#p = (p')/a = -4/2 = -2# and #q = (q')/a = 7/2#

Factored form: f(x) = 2(x - 2)(x + 7/2) = (x - 2)(2x + 7)