# How do you simplify the expression (b/(b+3)+2)/(b^2-2b-8)?

Apr 4, 2017

$\frac{3}{{b}^{2} - b - 12}$

#### Explanation:

Given:$\text{ } \frac{\frac{b}{b + 3} + 2}{{b}^{2} - 2 b - 8}$

$\textcolor{b l u e}{\text{Consider the numerator: } \frac{b}{b + 3} + \frac{2}{1}}$

I have deliberately chosen to write 2 as $\frac{2}{1}$. Although correct this is not normally done.

$\textcolor{g r e e n}{\left[\frac{b}{b + 3}\right] + \left[\frac{2}{1} \textcolor{red}{\times 1}\right]}$

$\textcolor{g r e e n}{\left[\frac{b}{b + 3}\right] + \left[\frac{2}{1} \textcolor{red}{\times \frac{b + 3}{b + 3}}\right]}$

$\left[\frac{b}{b + 3}\right] + \left[\frac{2 b + 6}{b + 3}\right]$

$\frac{3 b + 6}{b + 3}$
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$\textcolor{b l u e}{\text{Consider the denominator: } {b}^{2} - 2 b - 8}$

Notice that $2 \times \left(- 4\right) = - 8 \mathmr{and} 2 - 4 = - 2$

Write as $\left(b - 4\right) \left(b + 2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

numerator$\div$ denominator$\to \frac{3 b + 6}{b + 3} \div \left(b - 4\right) \left(b + 2\right)$

$\text{ "=" } \frac{3 b + 6}{b + 3} \times \frac{1}{\left(b - 4\right) \left(b + 2\right)}$

Factor out the 3 from $3 b + 6 \to 3 \left(b + 2\right)$

$\text{ "=" } \frac{3 \cancel{\left(b + 2\right)}}{b + 3} \times \frac{1}{\left(b - 4\right) \cancel{\left(b + 2\right)}}$

$\text{ "=" } \frac{3}{{b}^{2} - b - 12}$