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

May 20, 2016

(3y + 1)(y - 5)

#### Explanation:

Use the new AC Method (Socratic Search)
$f \left(y\right) = 3 {y}^{2} - 14 y - 5 =$ 3(x + p)(x + q)
Converted trinomial $f ' \left(y\right) = {y}^{2} - 14 y - 15 =$ (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), $p = \frac{p '}{a} = \frac{1}{3}$and $q = \frac{q '}{a} = - \frac{15}{3} = - 5$.
Factored form $f \left(y\right) = 3 \left(y + \frac{1}{3}\right) \left(y - 5\right) = \left(3 y + 1\right) \left(y - 5\right)$