# How do you solve 3x^2+5x+2=0 by factoring?

Mar 27, 2015

Assuming $3 {x}^{2} + 5 x + 2$ can be factored into
$\left(a x + b\right) \left(c x + d\right)$ with integer values for $a , b , c , d$
and noting that all of $a , b , c , d$ will be non-negative (from the form of the original quadratic)
we only have the possibilities
$\left\{a , c\right\} = \left\{1 , 3\right\}$ and
$\left\{b , d\right\} = \left\{1 , 2\right\}$

There are only 2 combinations to attempt:
$\left(1 x + 1\right) \left(3 x + 2\right)$
and
$\left(1 x + 2\right) \left(3 x + 1\right)$

revealing that $3 {x}^{2} + 5 x + 2 = 0$ can be factored as
$\left(x + 1\right) \left(3 x + 2\right) = 0$

So either
$x + 1 = 0$ which implies $x = - 1$
or
$3 x + 2 = 0$ which implies $x = - \frac{2}{3}$