Question

How do you combine `(2)/(x) + (2)/(x-1) - (2)/(x-2)`?

Answer 1

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

Explanation:

This could get messy.
All of the denominators need to be the same, and the only way we can change them is by multiplying the term by a special form of `1`.

So, to get them to match we will need all of them to have this in the denominator: `(x)(x-1)(x-2)`.

NOTE: Don't bother to FOIL anything out yet...

The first term will be multiplied by `((x-1)(x-2))/((x-1)(x-2))` to give
`(2(x-1)(x-2))/(x(x-1)(x-2))`

The second term will be multiplied by `((x)(x-2))/((x)(x-2))` to give
`(2(x)(x-2))/(x(x-1)(x-2))`

The third term will be multiplied by `((x)(x-1))/((x)(x-1))` to give
`(2(x)(x-1))/(x(x-1)(x-2))`

Now to put it all together...
`(2(x-1)(x-2))/(x(x-1)(x-2)) + (2(x)(x-2))/(x(x-1)(x-2)) -(2(x)(x-1))/(x(x-1)(x-2))`

With common denominators they can be simply combined like so...

`((2(x-1)(x-2)) + (2(x)(x-2)) -(2(x)(x-1)))/(x(x-1)(x-2))`

Now we need to FOIL stuff out...

`((2(x^2-3x-2)) + (2(x^2-2x)) -(2(x^2-x)))/(x^3-3x^2-2x)`

`((2x^2-6x-4) + (2x^2-4x) -(2x^2-2x))/(x^3-3x^2-2x)`

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