# How do you write the partial fraction decomposition of the rational expression (5x-1)/(x^2-x-2)?

Jan 4, 2016

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

#### Explanation:

General form to break into partial fractions is
$\frac{a x + b}{c {x}^{2} + \mathrm{dx} + e} = \frac{A}{x + C} + \frac{B}{x + D}$

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

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

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

we can equate the coefficients of numerators of both sides
$A + B = 5 , A - 2 B = - 1$
Subtracting both gives $3 B = 6$
or B = 2
And A=3
thus

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