# How do you factor completely 3x^2 - 5x - 2?

Jun 23, 2016

$3 {x}^{2} - 5 x - 2 = \textcolor{b l u e}{\left(1 x - 2\right) \left(3 x + 1\right)}$

#### Explanation:

Looking for $\left\{a , b , c , d\right\}$ for a factoring $\left(a x + c\right) \left(b x + d\right) = 3 {x}^{2} - 5 x - 2$

We see that
$\textcolor{w h i t e}{\text{XXX}} a b = 3$
$\textcolor{w h i t e}{\text{XXX}} c d = - 2$
and
$\textcolor{w h i t e}{\text{XXX}} a d + c b = - 5$

In the hopes of finding integer values for $\left\{a , b , c , d\right\}$
we note that the factors of $3$ are $\left\{1 , 3\right\}$
so if $a , b$ are integers $\left\{a , b\right\} = \left\{1 , 3\right\}$
and
we also note the the factors of $- 2$ are $\left\{\begin{matrix}- 1 & 2 \\ 1 & - 2\end{matrix}\right\}$

We can test these possible combinations looking for $a d + c b = 5$

{: (underline(a),underline(b),underline(c),underline(d),,underline(ad+cb)), (1,3,-1,2,,-1), (1,3,2,-1,,+5), (1,3,1,-2,,+1), (color(red)(1),color(red)(3),color(red)(-2),color(red)(1),,color(red)(-5)) :}

We have found values that match our requirements:
$\textcolor{w h i t e}{\text{XXX}} < a , b , c , d > = < 1 , 3 , - 2 , 1 >$
So
$\textcolor{w h i t e}{\text{XXX}} \left(a x + c\right) \left(b x + d\right) = \left(1 x - 2\right) \left(3 x + 1\right)$