# How do you simplify and find the restrictions for (x^3-2x^2-8x)/(x^2-4x)?

Sep 17, 2016

$\frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x} = x + 2$, but $x$ cannot take values $x = 0$ and $x = 4$

#### Explanation:

$\frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x}$

= $\frac{x \left({x}^{2} - 2 x - 8\right)}{x \left(x - 4\right)}$

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

= $\frac{x \left(x \left(x - 4\right) + 2 \left(x - 4\right)\right)}{x \left(x - 4\right)}$

= $\frac{x \left(x - 4\right) \left(x + 2\right)}{x \left(x - 4\right)}$

= $\frac{\cancel{x} \cancel{\left(x - 4\right)} \left(x + 2\right)}{1 \cancel{x} \cancel{\left(x - 4\right)}}$

= $x + 2$

but $x$ cannot take values $x = 0$ and $x = 4$ as that makes the denominator ${x}^{2} - 4 x = 0$.

Sep 17, 2016

$\frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x} \equiv \left(x + 2\right)$

#### Explanation:

Given:$\text{ } \frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x}$

The equation becomes 'undefined' if the denominator is 0. You are 'not allowed' to divide by 0.

$\textcolor{b l u e}{\text{To determine the excluded value}}$

Set $0 = {x}^{2} - 4 x \text{ " ->" } 0 = x \left(x - 4\right)$

Thus the excluded values for x are: $x = 0 \text{ and } x = 4$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Simplifying the expression}}$

Factor out $x$ for top and bottom (numerator and denominator)

$\implies \frac{{\cancel{x}}^{1} \left({x}^{2} - 2 x - 8\right)}{{\cancel{x}}^{1} \left(x - 4\right)} \text{ "=" } \frac{{x}^{2} - 2 x - 8}{x - 4}$

Factorizing the top

$\implies \frac{\left(x + 2\right) \left(x - 4\right)}{x - 4} = \left(x + 2\right) \times \frac{x - 4}{x - 4}$

But $\frac{x - 4}{x - 4} = 1$

Thus

$\text{ } \textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \frac{{x}^{3} - 2 {x}^{2} - 8 x}{{x}^{2} - 4 x} \equiv \left(x + 2\right) \textcolor{w h i t e}{\frac{2}{2}} |}}}$