How do you factor #3y^2-10y-8=0#?

1 Answer
Apr 12, 2016

Answer:

(3y + 2)(y - 4)

Explanation:

Use the new AC Method (Socratic Search)
f(y) = 3y^2 - 10y - 8 = 3(y + p)(y + q)
Converted trinomial: f'(y) = y^2 - 10y - 24 = (y + p'(y + q')
p' and q' have opposite sign because ac < 0.
Factor pairs of (-24) --> (-2, 12)(2, -12) This sum is -10 = b.
Then, p' = 2 and q' = -10.
Back to f9y): p = (p')/a = 2/3 and q = (q')/a = -12/3 = - 4.
Factored form --> f(y) = 3(y + 2/3)(y - 4) = (3y + 2)(y - 4)