How do you factor #3y^2 - 14y - 5#?

1 Answer
May 20, 2016

Answer:

(3y + 1)(y - 5)

Explanation:

Use the new AC Method (Socratic Search)
#f(y) = 3y^2 - 14y - 5 =# 3(x + p)(x + q)
Converted trinomial #f'(y) = y^2 - 14y - 15 =# (x + p')(x + q')
p' and q' have opposite signs because ac < 0.
Factor pairs of (ac = -15) --> (-1, 15)(1, -15). This sum is (-14 = b). Then p' = 1 and q' = -15.
Back to original trinomial f(y), #p = (p')/a = 1/3 #and #q = (q')/a = -15/3 = -5#.
Factored form #f(y) = 3(y + 1/3)(y - 5) = (3y + 1)(y - 5)#