# How do you factor 3y^2+7y-6?

Aug 6, 2015

$3 {y}^{2} + 7 y - 6 = \left(3 y - 2\right) \left(y + 3\right)$

#### Explanation:

You could use the quadratic formula for this
but (hoping for integer coefficient factors):

Since $\left(a y + b\right) \left(c y + d\right) = \left(a c\right) {y}^{2} + \left(a d + b c\right) y + b d$

and for the given example
$\textcolor{w h i t e}{\text{XXXX}}$(ac) = 3$\mathmr{and}$(bd) = (-6)

we want to find $\left(a , c\right)$, factors of 3
$\textcolor{w h i t e}{\text{XXXX}}$only obvious factors are [(1,3), (3,1)]
and factors, $\left(b , d\right)$,of (-6)
$\textcolor{w h i t e}{\text{XXXX}}$[(1,-6), (-1,6), (2,-3), (-2,3), (6, -1), (-6,1), (3,-2), (-3,2)]
which satisfy
$\textcolor{w h i t e}{\text{XXXX}}$(ad+bc) = 7

with a small amount of effort we find
$\textcolor{w h i t e}{\text{XXXX}}$$\left(a , c\right) = \left(3 , 1\right)$ and $\left(b , d\right) = \left(- 2 , 3\right)$