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

Nov 8, 2015

$\frac{2 {x}^{2} - 8 x - 4}{{x}^{3} - 3 {x}^{2} - 2 x}$

#### 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: $\left(x\right) \left(x - 1\right) \left(x - 2\right)$.

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

The first term will be multiplied by $\frac{\left(x - 1\right) \left(x - 2\right)}{\left(x - 1\right) \left(x - 2\right)}$ to give
$\frac{2 \left(x - 1\right) \left(x - 2\right)}{x \left(x - 1\right) \left(x - 2\right)}$

The second term will be multiplied by $\frac{\left(x\right) \left(x - 2\right)}{\left(x\right) \left(x - 2\right)}$ to give
$\frac{2 \left(x\right) \left(x - 2\right)}{x \left(x - 1\right) \left(x - 2\right)}$

The third term will be multiplied by $\frac{\left(x\right) \left(x - 1\right)}{\left(x\right) \left(x - 1\right)}$ to give
$\frac{2 \left(x\right) \left(x - 1\right)}{x \left(x - 1\right) \left(x - 2\right)}$

Now to put it all together...
$\frac{2 \left(x - 1\right) \left(x - 2\right)}{x \left(x - 1\right) \left(x - 2\right)} + \frac{2 \left(x\right) \left(x - 2\right)}{x \left(x - 1\right) \left(x - 2\right)} - \frac{2 \left(x\right) \left(x - 1\right)}{x \left(x - 1\right) \left(x - 2\right)}$

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

$\frac{\left(2 \left(x - 1\right) \left(x - 2\right)\right) + \left(2 \left(x\right) \left(x - 2\right)\right) - \left(2 \left(x\right) \left(x - 1\right)\right)}{x \left(x - 1\right) \left(x - 2\right)}$

Now we need to FOIL stuff out...

$\frac{\left(2 \left({x}^{2} - 3 x - 2\right)\right) + \left(2 \left({x}^{2} - 2 x\right)\right) - \left(2 \left({x}^{2} - x\right)\right)}{{x}^{3} - 3 {x}^{2} - 2 x}$

$\frac{\left(2 {x}^{2} - 6 x - 4\right) + \left(2 {x}^{2} - 4 x\right) - \left(2 {x}^{2} - 2 x\right)}{{x}^{3} - 3 {x}^{2} - 2 x}$

$\frac{2 {x}^{2} - 8 x - 4}{{x}^{3} - 3 {x}^{2} - 2 x}$