How do you express as a partial fraction #(5x-1)/(x^2-x-2)#?

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Answer 1 · 2015-07-18T16:05:34

#(5x-1)/(x^2-x-2) = 3/(x-2)+2/(x+1)#

#(5x-1)/(x^2-x-2)#

The denominator can be factored using real numbers, so we do that first:

#x^2-x-2 = (x-2)(x+1)#

Now we want:

#A/(x-2) + B/(x+1) = (5x-1)/(x^2-x-2)#

One way of proceeding is to get a single fraction on the left:

#A/(x-2) + B/(x+1) = (A(x+1)+B(x-2))/((x-2)(x+1))#

# = ((Ax+Bx)+(A-2B))/(x^2-x-2)#

So we want:

#((A+B)x+(A-2B))/(x^2-x-2) = (5x-1)/(x^2-x-2)#

That means we need:

#A+B# #" " = # #" "## 5#
#A-2B##" "# # =# #-1#

Subtract the second from the first (change the signs and add) to get:

#3B# # = # #6#

So #B=2# and in order to get #A+B=5# we must also have #A=3#

We have:

#3/(x-2)+2/(x+1)#

It is worth taking a moment to check our answer:

#(overbrace(3(x+1))^(3x+3)+overbrace(2(x-2))^(2x-4))/((x-2)(x+1))#.

Yes the top simplifies to #5x-1#, so we should be OK.