# How do you simplify 4/(x-2)-3/(x+1)+2/(x^2-x-2)?

$\frac{x + 12}{\left(x - 2\right) \left(x + 1\right)} = \frac{x + 12}{{x}^{2} - x - 2}$

#### Explanation:

$\frac{4}{x - 2} - \frac{3}{x + 1} + \frac{2}{{x}^{2} - x - 2}$

We need the denominators to be the same. We can do that by multiplying through with various forms of the number 1:

$\frac{4}{x - 2} \left(1\right) - \frac{3}{x + 1} \left(1\right) + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{4}{x - 2} \left(\frac{x + 1}{x + 1}\right) - \frac{3}{x + 1} \left(\frac{x - 2}{x - 2}\right) + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{4 \left(x + 1\right)}{\left(x - 2\right) \left(x + 1\right)} - \frac{3 \left(x - 2\right)}{\left(x + 1\right) \left(x - 2\right)} + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{4 x + 4}{\left(x - 2\right) \left(x + 1\right)} - \frac{3 x - 6}{\left(x + 1\right) \left(x - 2\right)} + \frac{2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{\left(4 x + 4\right) - \left(3 x - 6\right) + 2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{4 x + 4 - 3 x + 6 + 2}{\left(x - 2\right) \left(x + 1\right)}$

$\frac{x + 12}{\left(x - 2\right) \left(x + 1\right)} = \frac{x + 12}{{x}^{2} - x - 2}$