How do you factor the expression #16y^2 - 12y + 2#?

1 Answer
Apr 26, 2016

f(y) = 2(4y - 1)(2y - 1)

Explanation:

There is a systematic, non- guessing, new AC Method (Socratic Search)
#f(y) = 16y^2 - 12y + 2 =# 16(y + p)(y + q)
Converted trinomial #f'(y) = x^2 - 12y + 32 =# (x + p')(x + q')
p' and q' have same sign, because ac > 0.
Factor pairs of (ac = 32) --> (-4, -8) . This sum is -12 = b. Then, p' = -4 and q' = -8.
Back to f(y), #p = (p')/a = -4/16 = -1/4# and #q = (q')/a = -8/16 = -1/2#.
Factored form: #f(y) = 16(y - 1/4)(y - 1/2) = 2(4y - 1)(2y - 1)#