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

1 Answer
Maggie
Mar 6, 2016

{-2*x-3}/{x^2-x}={-5}/{x-1}+3/x

Explanation:

We begin with
{-2*x-3}/{x^2-x}
First we factor the bottom to get
{-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
A/{x-1}+B/x, where A and B are real numbers.

Starting with
A/{x-1}+B/x, we use fraction addition rules to get
{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

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

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
{-5}/{x-1}+3/x={-2*x-3}/{x^2-x}

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