Question

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

Answer 1

`x!=0 and x!= 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 - 4x=0`?

`x(x-4) = 0" "larr` factorise

`x = 0 " or " x = 4`

These are the restrictions.