# How do you simplify ((j^2-16)/(j^2+10j+16))/(15/(j+8))?

Dec 18, 2016

$= \left(\frac{1}{15}\right) \left(j - 2\right) \left(j + 4\right)$

$= \frac{1}{15} {j}^{2} + \frac{2}{15} j - \frac{8}{15}$

#### Explanation:

$\frac{\left(\frac{{j}^{2} - 16}{{j}^{2} + 10 j + 16}\right)}{\left(\frac{15}{j + 8}\right)}$

Rather than dividing two fractions, use the fact that dividing is the same as multiplying by the reciprocal.
$= \left(\frac{{j}^{2} - 16}{{j}^{2} + 10 j + 16}\right) \cdot \left(\frac{j + 8}{15}\right)$

$= \frac{\left({j}^{2} - 16\right) \left(j + 8\right)}{\left({j}^{2} + 10 j + 16\right) \left(15\right)}$

Now factor the terms on the numerator and denominator:
$= \frac{\left(j - 4\right) \left(j + 4\right) \left(j + 8\right)}{\left(j + 8\right) \left(j + 2\right) \left(15\right)}$

$= \frac{\left(j - 2\right) \left(j + 2\right) \left(j + 4\right) \left(j + 8\right)}{\left(j + 8\right) \left(j + 2\right) \left(15\right)}$

Cancel terms that occur on both the numerator and denominator:
$= \frac{\left(j - 2\right) \cancel{\left(j + 2\right)} \left(j + 4\right) \cancel{\left(j + 8\right)}}{\cancel{\left(j + 8\right)} \cancel{\left(j + 2\right)} \left(15\right)}$

$= \frac{\left(j - 2\right) \left(j + 4\right)}{15}$
$= \left(\frac{1}{15}\right) \left(j - 2\right) \left(j + 4\right)$
$= \left(\frac{1}{15}\right) \left({j}^{2} + 2 j - 8\right)$