Question

How do you simplify `2/x + 1/(x-2) + 3/(x-2)^2`?

Answer 1

`(3x^2 - 7x + 8)/(x(x-2)^2)`

Explanation:

Your starting expressiin is

`2/x + 1/(x-2) + 3/(x-2)^2`

Your goal here is to find the common denominator for these three fractions and multiply each of them so that they all have this common denominator.

Once you do that, you can add their numerators like you would any three expressions.

So, your three denominators are

`x" "`, `" "x-2" "`, and `" "(x-2)^2 = (x-2)(x-2)`

Notice that the first denominator is missing `(x-2)^2`, the second an `x(x-2)`, and the third an `x`.

The common denominator will be `x(x-2)^2`. To get each fraction to have this denominator, you need to multiply the fist one by `1 = (x-2)^2/(x-2)^2`, the second one by `1 = (x(x-2))/(x(x-2))`, and the third one by `1 = x/x`.

This will get you

`2/x * (x-2)^2/(x-2)^2 + 1/(x-2) * (x(x-2))/(x(x-2)) + 3/(x-2)^2 * x/x`

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

Now you can focus on the numerators. Expand the parantheses to get

`2 * (x^2 - 4x + 4) + x^2 - 2x + 3x`

`2x^2 - 8x + 8 +x^2 + x = 3x^2 - 7x + 8`

The final version of the expression will be

`(3x^2 - 7x + 8)/(x(x-2)^2)`