# How do you simplify (x^2+6x-27)div(3x^2+27x)/(x+5)?

Sep 19, 2016

If $x \ne - 9$ and $x \ne 3$ then

$\left({x}^{2} + 6 x - 27\right) \div \frac{3 {x}^{2} + 27 x}{x + 5} = \textcolor{g r e e n}{\frac{{x}^{2} + 2 x - 15}{3 x}}$

#### Explanation:

Note that
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 6 x - 27 = \left(x + 9\right) \left(x - 3\right)$
and that
$\textcolor{w h i t e}{\text{XXX}} 3 {x}^{x} + 27 = 3 x \left(x + 9\right)$

and since $q \div \frac{a}{b} \Rightarrow q \cdot \frac{b}{a}$

$\left({x}^{2} + 6 x - 27\right) \div \frac{3 {x}^{2} + 27 x}{x + 5}$

$\textcolor{w h i t e}{\text{XXX")=(cancel(x+9))(x-3)xx(x+5)/(3x(cancel(x+9)))color(white)("XXX}}$
$\textcolor{w h i t e}{\text{XXX}}$provided $x + 9 \ne 0 \mathmr{and} 3 x \ne 0$

$\textcolor{w h i t e}{\text{XXX}} = \frac{{x}^{2} + 2 x - 15}{3 x}$

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Note that if $x = - 9$ or $x = 3$
then $3 {x}^{2} + 27 x = 0$ and the division is undefined