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

Feb 9, 2017

$x \ne 0 \mathmr{and} x \ne 4$
These values will make the denominator 0.

#### Explanation:

The restrictions in this case will be any value(s) which make(s) the denominator equal to 0.

You can have 0 as a numerator, but not as a denominator.

What will make ${x}^{2} - 4 x = 0$?

$x \left(x - 4\right) = 0 \text{ } \leftarrow$ factorise

$x = 0 \text{ or } x = 4$

These are the restrictions.