# How do you solve 3x^2-13x-10=0 by factoring?

Oct 10, 2015

$\left(3 x + 2\right) \left(x - 5\right)$

#### Explanation:

Because $3$ can only have factors of $3$ and 1
you know those will be your coefficients for $x$.
That is your factors will be of the form
$\textcolor{w h i t e}{\text{XXX}} \left(3 x + a\right) \left(x + b\right)$

You also know that one of $a$ or $b$ will have $+$ and one side will have $-$ due to the $\textcolor{red}{-} 10$.

The options for $10$ are
$2 \times 5$ or $1 x 10$.

Using $2 \times 5$
if you multiply the $5$ by $3$ it gives you $15$
and the $2$ by $1$1 you get $2$.

Because $13$ in the center (of the original form) is negative
you know the $5$ needs to be negative

and the factors are
$\textcolor{w h i t e}{\text{XXX}} \left(3 x + 2\right) \left(x - 5\right)$