How do you factor #16x^2 + 32x + 7 = 0#?

1 Answer
Jun 3, 2015

#f(x) = 16x^2 + 32x + 7 #= 16(x - p)(x - q). I use the new AC Method.
Convert f(x) to# f'(x) = x^2 + 32x + 112 =# (x - p')(x - q').
Compose factor pairs of (a.c = 112)--> (2, 56)(4, 28). This sum = 32 = b. Then p' = 4 and q' = 28.
back to f(x), #p = (p')/a = 4/16 = 1/4#, and #q = (q')/a = 28/16 = 7/4.#

Factored form:# f(x) = 16(x + 1/4)(x + 7/4) = (4x + 1)(4x + 7)#