# How do you express (5x-1)/(x^2-x-2) in partial fractions?

Nov 29, 2016

Factorize the denominator, then calculate the numerators of partial fractions.

#### Explanation:

First you find the zeroes of the denominator to factorize it.

${x}^{2} - x - 2 = 0$

$x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}$

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

Now pose:

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

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

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

Equating the coefficient of the same order in the numerators:

$A + B = 5$
$A - 2 B = - 1$

and solving the system:

$A = 3$
$B = 2$

thus:

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$