How do you factor #16x^2 - 28x - 8#?

1 Answer
Sep 27, 2015

Factor #y = 16x^2 - 28x - 8#

Explanation:

#f(x) = 4y = 4(4x^2 - 7x - 2)#
Factor #y = 4x^2 - 7x - 2 =# 4(x + p)(x + q)
I use the new AC Method to factor trinomials (Socratic Search).
Converted trinomial #y' = x^2 - 7x - 8.#
Factor pairs of (-8) --> (-1, 8). This sum is 7 = -b. Then p' = 1 and q' = -8
Therefor #p = 1/4# and #q = -8/4 = - 2#
Factored form of #y = 4(x + 1/4)(x - 2) = (4x + 1)(x - 2).#
Finally:
#f(x) = 4(4x + 1)(x - 2)#