# How do you express (-2x-3)/(x^2-x)  in partial fractions?

Mar 6, 2016

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

#### Explanation:

We begin with
$\frac{- 2 \cdot x - 3}{{x}^{2} - x}$
First we factor the bottom to get
$\frac{- 2 \cdot x - 3}{x \left(x - 1\right)}$.

We have a quadratic on the bottom and a linear on the top this means we are looking for something of the form
$\frac{A}{x - 1} + \frac{B}{x}$, where $A$ and $B$ are real numbers.

Starting with
$\frac{A}{x - 1} + \frac{B}{x}$, we use fraction addition rules to get
$\frac{A \cdot x}{x \left(x - 1\right)} + \frac{B \cdot \left(x - 1\right)}{x \left(x - 1\right)} = \frac{A \cdot x + B x - B}{x \left(x - 1\right)}$

We set this equal to our equation

$\frac{\left(A + B\right) x - B}{x \left(x - 1\right)} = \frac{- 2 \cdot x - 3}{x \left(x - 1\right)}$.

From this we can see that
$A + B = - 2$ and $- B = - 3$.

We end up with
$B = 3$ and $A + 3 = - 2$ or $A = - 5$.

So we have
$\frac{- 5}{x - 1} + \frac{3}{x} = \frac{- 2 \cdot x - 3}{{x}^{2} - x}$