How do you factor #4x^2+4x-15#?

1 Answer
Aug 6, 2015

Answer:

Factor# y = 4x^2 + 4x - 15#

Explanation:

#y = 4x^2 + 4x - 15 =# 4(x - p)(x - q)
I use the new AC Method to factor trinomials.
Converted trinomial #y' = x^2 + 4x - 60 =# (x - p')(x - q')
Factor pairs of (-60) --> ...(-4, 15)(-5, 12)(-6, 10). This sum is 4 = b. Then p' = -6 and q' = 10.
Therefor, #p = (p')/a = -6/4 = -3/2# and #q = (q')/a = 10/4 = 5/2.#
Factored form: #y = 4(x - 3/2)(x - 5/2) = (2x - 3)(2x + 5)#

NOTE. This new AC Method can avoid the lengthy factoring by grouping