# How do you factor 3y^2+7y-20?

$f \left(y\right) = 3 {y}^{2} + 7 y - 20$= 3(x - p)(x - q). I use the new AC method
Converted $f ' \left(y\right) = {y}^{2} + 7 y - 60.$ a and c have different signs. Compose factor pairs of (a.c = -60) --> (-3, 20)(-4, 15)(-5, 12). This sum is 7 = b. Then p' = -5 and q' = 12.
Back to f(y) -> $p = - \frac{5}{3} \mathmr{and} q = \frac{12}{3} = 4.$
Factored form $f \left(y\right) = 3 \left(y - \frac{5}{3}\right) \left(y + 4\right) = \left(3 y - 5\right) \left(y + 4\right) .$
Check by developing: $f \left(y\right) = 3 {y}^{2} + 12 y - 5 y - 20.$.OK