Question

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

Answer 1

`(x^3-2x^2-8x)/(x^2-4x)=x+2`, but `x` cannot take values `x=0` and `x=4`

Explanation:

`(x^3-2x^2-8x)/(x^2-4x)`

= `(x(x^2-2x-8))/(x(x-4))`

= `(x(x^2-4x+2x-8))/(x(x-4))`

= `(x(x(x-4)+2(x-4)))/(x(x-4))`

= `(x(x-4)(x+2))/(x(x-4))`

= `(cancel(x)cancel((x-4))(x+2))/(1cancel(x)cancel((x-4)))`

= `x+2`

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

Answer 2

`(x^3-2x^2-8x)/(x^2-4x) -= (x+2)`

Explanation:

Given:`" "(x^3-2x^2-8x)/(x^2-4x)`

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

`color(blue)("To determine the excluded value")`

Set `0=x^2-4x" " ->" " 0=x(x-4)`

Thus the excluded values for x are: `x=0" and "x=4`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Simplifying the expression")`

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

`=>(cancel(x)^1(x^2-2x-8))/(cancel(x)^1(x-4))" "=" "(x^2-2x-8)/(x-4)`

Factorizing the top

`=>((x+2)(x-4))/(x-4) = (x+2)xx(x-4)/(x-4)`

But `(x-4)/(x-4)=1`

Thus

`" "color(blue)(bar(ul(|color(white)(2/2)(x^3-2x^2-8x)/(x^2-4x) -= (x+2)color(white)(2/2)|)))`