# How do you factor 15v^2-19v-10?

Jun 5, 2017

$15 {v}^{2} - 19 v - 10 = \left(5 v + 2\right) \left(3 v - 5\right)$

#### Explanation:

Given:

$15 {v}^{2} - 19 v - 10$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c$

where $a = 15$, $b = - 19$ and $c = - 10$

That is:

$\Delta = {\left(\textcolor{b l u e}{- 19}\right)}^{2} - 4 \left(\textcolor{b l u e}{15}\right) \left(\textcolor{b l u e}{- 10}\right)$

$\textcolor{w h i t e}{\Delta} = 361 + 600$

$\textcolor{w h i t e}{\Delta} = 961$

$\textcolor{w h i t e}{\Delta} = {31}^{2}$

Since $\Delta > 0$ is a perfect square, the given quadratic will factor perfectly using integers.

Let's use an AC method.

Look for a pair of factors of $A C = 15 \cdot 10 = 150$ which differ by $B = 19$. (We look for a pair of factors with this as a difference rather than a sum, since the sign of the constant term is negative.)

The pair $25 , 6$ works in that $25 \cdot 6 = 150$ and $25 - 6 = 19$

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

$15 {v}^{2} - 19 v - 10 = \left(15 {v}^{2} - 25 v\right) + \left(6 v - 10\right)$

$\textcolor{w h i t e}{15 {v}^{2} - 19 v - 10} = 5 v \left(3 v - 5\right) + 2 \left(3 v - 5\right)$

$\textcolor{w h i t e}{15 {v}^{2} - 19 v - 10} = \left(5 v + 2\right) \left(3 v - 5\right)$