How do you factor #15y^2 + 26y +8=0#?

1 Answer
May 9, 2016

(5y + 2)(3y + 4)

Explanation:

Use the new AC Method to factor trinomials (Socratic Search).
#f(y) = 15y^2 + 26y + 8 = #15(y + p)(y + q).
Converted trinomial #f'(y) = y^2 + 26y + 120 =# (y + p')(y + q').
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 120) --> ...(#5, 24)(6, 20).# This sum is (26 = b). Therefor, p' = 6 and q' = 20.
Back to original trinomial y --> #p = (p')/a = 6/15 = 2/5# and
#q = (q')/a = 20/15 = 4/3#
Factored for --> f(y) = 15(y + 2/5)(y + 4/3) = (5y + 2)(3y + 4)