# How do you decompose (x-6)/(x^2-2x) into partial fractions?

Dec 21, 2016

Begin with this equation:

$\frac{x - 6}{{x}^{2} - 2 x} = \frac{A}{x} + \frac{B}{x - 2}$

#### Explanation:

Multiply both sides of the equation by $x \left(x - 2\right)$:

$x - 6 = A \left(x - 2\right) + B x$

Let x = 0:

$0 - 6 = A \left(0 - 2\right) + B \left(0\right)$

$- 6 = A \left(- 2\right)$

$A = 3$

Let x = 2:

$2 - 6 = A \left(2 - 2\right) + B \left(2\right)$

$- 4 = 2 B$

$B = - 2$

$A = 3 \mathmr{and} B = - 2$

Check:

$\frac{3}{x} - \frac{2}{x - 2} =$

$\frac{3}{x} \frac{x - 2}{x - 2} - \frac{2}{x - 2} \frac{x}{x} =$

$\frac{3 x - 6}{x \left(x - 2\right)} - \frac{2 x}{x \left(x - 2\right)} =$

$\frac{3 x - 6 - 2 x}{x \left(x - 2\right)} =$

$\frac{x - 6}{{x}^{2} - 2 x}$

The checks.

$\frac{x - 6}{{x}^{2} - 2 x} = \frac{3}{x} - \frac{2}{x - 2}$