How do you factor #(7x^2+11x-6)/(x^2-4)#?

1 Answer
Jul 21, 2016

Answer:

#(7x - 3)/(x - 2)#

Explanation:

First, factor the numerator y by the new AC Method (Socratic Search).
y = 7x^2 + 11x - 6 = 7(x + p)(x + q)
Converted trinomial y' = x^2 + 11x - 42 = (x + p')(x + q')
p' and q' have opposite signs. Factor pairs of (ac = -42) -->
(-2, 21(-3, 14). This last sum is (11 = b). Then p' = -3 and q' = 14.
Back to original trinomial y, #p = p'/a = -3/7#, and
#q = (q')/a = 14/7 = 2.#
Factor form of y:
y = 7(x - 3/7)(x + 2) = (7x - 3)(x + 2)# Factored form of #f(x) = ((7x - 3)(x + 2))/((x - 2)(x + 2)) =# #f(x) = (7x - 3)/(x - 2)#