# How do you simplify (v^2 + v - 12)/ (v^2 + 6v + 8) div (2v - 6)/( v +2)?

Jul 6, 2016

$\frac{1}{2}$

#### Explanation:

the trynomial

${v}^{2} + \left({v}_{1} + {v}_{2}\right) v + \left({v}_{1} \cdot {v}_{2}\right) v$

is obtained by the product:

$\left(v + {v}_{1}\right) \left(v + {v}_{2}\right)$

so

${v}^{2} + v - 12 = \left(v + 4\right) \left(v - 3\right)$

and

${v}^{2} + 6 v + 8 = \left(v + 4\right) \left(v + 2\right)$

So you can rewrite

$\frac{{v}^{2} + v - 12}{{v}^{2} + 6 v + 8} \div \frac{2 v - 6}{v + 2} = \frac{\cancel{\left(v + 4\right)} \left(v - 3\right)}{\cancel{\left(v + 4\right)} \left(v + 2\right)} \div \frac{2 \left(v - 3\right)}{v + 2}$

$= \frac{\cancel{v - 3}}{\cancel{v + 2}} \cdot \frac{\cancel{v + 2}}{2 \cancel{\left(v - 3\right)}} = \frac{1}{2}$