# How do you factor 2x^2-x-21?

Aug 10, 2015

$2 {x}^{2} - x - 21 = \left(2 x - 7\right) \left(x + 3\right)$

#### Explanation:

Assuming that $2 {x}^{2} - x - 21$
can be written as $\left(a x \pm c\right) \left(b x \pm d\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$with $a , b , c , d \in \mathbb{Z}$

Since the sign of the last term ($- 21$) is negative the interior signs of the two binomials must be opposite
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow$ we want $\left(a x - c\right) \left(b x + d\right)$;
and
since the sign of the middle term ($- x$) is negative
$c b > a d$

Looking at factors of 2: $\left(a b\right) \in \left\{\left(2 , 1\right)\right\}$
and factors of 21: $\left(c , d\right) \in \left\{\begin{matrix}3 & 7 \\ 7 & 3 \\ 1 & 21 \\ 21 & 1\end{matrix}\right\}$

we find
$\textcolor{w h i t e}{\text{XXXX}}$$\left(a , b\right) = \left(2 , 1\right)$ and $\left(c , d\right) = \left(7 , 3\right)$
give us
$\textcolor{w h i t e}{\text{XXXX}}$$a \cdot d - b \cdot c = \left(- 1\right)$

So our factors are
$\textcolor{w h i t e}{\text{XXXX}}$$\left(2 x - 7\right) \left(x + 3\right)$