# How do you express as a partial fraction (5x-1)/(x^2-x-2)?

Jul 18, 2015

$\frac{5 x - 1}{{x}^{2} - x - 2} = \frac{3}{x - 2} + \frac{2}{x + 1}$

#### Explanation:

$\frac{5 x - 1}{{x}^{2} - x - 2}$

The denominator can be factored using real numbers, so we do that first:

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

Now we want:

$\frac{A}{x - 2} + \frac{B}{x + 1} = \frac{5 x - 1}{{x}^{2} - x - 2}$

One way of proceeding is to get a single fraction on the left:

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

$= \frac{\left(A x + B x\right) + \left(A - 2 B\right)}{{x}^{2} - x - 2}$

So we want:

$\frac{\left(A + B\right) x + \left(A - 2 B\right)}{{x}^{2} - x - 2} = \frac{5 x - 1}{{x}^{2} - x - 2}$

That means we need:

$A + B$ $\text{ } =$ $\text{ }$$5$
$A - 2 B$$\text{ }$ $=$ $- 1$

Subtract the second from the first (change the signs and add) to get:

$3 B$ $=$ $6$

So $B = 2$ and in order to get $A + B = 5$ we must also have $A = 3$

We have:

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

It is worth taking a moment to check our answer:

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

Yes the top simplifies to $5 x - 1$, so we should be OK.