# How do you factor 2v^2-4-7v ?

Mar 1, 2017

$2 {v}^{2} - 4 - 7 v = \left(2 v + 1\right) \left(v - 4\right)$

#### Explanation:

First put in standard order of decreasing powers of $v$:

$2 {v}^{2} - 7 v - 4$

Next, use an AC method:

Find a pair of factors of $A C = 2 \cdot 4 = 8$ which differ by $B = 7$.

The pair $8 , 1$ works.

Use this pair to split the middle term and factor by grouping:

$2 {v}^{2} - 7 v - 4 = 2 {v}^{2} - 8 v + v - 4$

$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = \left(2 {v}^{2} - 8 v\right) + \left(v - 4\right)$

$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = 2 v \left(v - 4\right) + 1 \left(v - 4\right)$

$\textcolor{w h i t e}{2 {v}^{2} - 7 v - 4} = \left(2 v + 1\right) \left(v - 4\right)$