How do you express $\frac{- 2 x - 3}{x^{2} - x}$ $(-2x-3)/(x^2-x)$ in partial fractions?

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How do you express $\frac{- 2 x - 3}{x^{2} - x}$ $(-2x-3)/(x^2-x)$ in partial fractions?

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Answer 1 · 2016-03-06T02:28:01

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

Explanation:

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

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

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

We set this equal to our equation

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

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

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

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

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How do you express $\frac{- 2 x - 3}{x^{2} - x}$ $(-2x-3)/(x^2-x)$ in partial fractions?

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How do you express $\frac{- 2 x - 3}{x^{2} - x}$ $(-2x-3)/(x^2-x)$ in partial fractions?