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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