How do you factor completely #3y^2 - 4y - 15 #?

1 Answer
Dec 13, 2015

Answer:

Factor #f(y) = 3y^2 - 4y - 15#

Ans: (3y + 5)(y - 3)

Explanation:

Use the new AC method to factor trinomials (Socratic Search)
#f(y) = 3y^2 - 4y - 15 =# 3(y + p )(y + q)
Converted trinomial #f'(y) = y^2 - 4y - 45 = #(y + p')( + q').
p' and q' have opposite signs. Factor pairs of (-45) --> (-3, 15)(-5, 9). This sum is 9 - 5 = 4 = -b. Then, the opposite sum gives: p' = 5 and q' = - 9.
Back to f(y), #p = (p')/a = 5/3# and #q = (q')/a = -9/3 = -3#.
Factored form: f(y) = 3(y + 5/3)(y - 3) = (3y + 5)(y - 3).