# Expand Into Partial Fractions

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• Nov 25th 2012, 10:42 PM
rrooggeerr
Expand Into Partial Fractions
I know how to do these problems but this one is giving me some trouble:
Expand into partial fractions

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

***Should I break it down into:
1) [A/(x+1)]+[B/(x-1)]+[C/(x+1)] or
2) [A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]

I tried it both ways and came up with different possibilities for the constraints:
If Broken Down Like #1 from Above
A+B+C=0
2B = 0
A+B-C=5

If Broken Down Like #2 from Above
A+C=0
B+2C=0
A+B-C=5
• Nov 25th 2012, 10:49 PM
MarkFL
Re: Expand Into Partial Fractions
Which way gives you the LCD that is the denominator of the original expression?
• Nov 25th 2012, 10:52 PM
rrooggeerr
Re: Expand Into Partial Fractions
I'm not sure because you can break down the original denominator (x+1)(x^2-1) into (x+1)(x+1)(x-1)
• Nov 25th 2012, 10:55 PM
MarkFL
Re: Expand Into Partial Fractions
Yes, and if you were to combine the 3 terms of both the choices you listed, which one would give you the correct common denominator?
• Nov 25th 2012, 10:59 PM
rrooggeerr
Re: Expand Into Partial Fractions
Option 1 would give me that..right?
• Nov 25th 2012, 11:07 PM
Prove It
Re: Expand Into Partial Fractions
No it wouldn't. Since two of them have \displaystyle \displaystyle \begin{align*} x + 1 \end{align*} as the denominator, the LCD is \displaystyle \displaystyle \begin{align*} (x + 1)(x - 1) \end{align*}, NOT \displaystyle \displaystyle \begin{align*} (x + 1)^2(x - 1) \end{align*} as required.

When you have repeated factors, you need one denominator to be \displaystyle \displaystyle \begin{align*} x + 1 \end{align*} and another to be \displaystyle \displaystyle \begin{align*} (x + 1)^2 \end{align*} in your partial fraction.
• Nov 25th 2012, 11:13 PM
rrooggeerr
Re: Expand Into Partial Fractions
so when I write it in the partial fractions I write it as:
[A/(x+1)]+[B/(x+1)^2]+[C/(x-1)]

??
• Nov 25th 2012, 11:16 PM
Prove It
Re: Expand Into Partial Fractions
Yes
• Nov 25th 2012, 11:20 PM
rrooggeerr
Re: Expand Into Partial Fractions
5/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)
multiply by LCD to get:

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

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

5= Ax^2 - A + Bx - B + Cx^2 + 2Cx + C

Group Common Terms:
5= x^2(A+C) + x(B+2C) + (-A-B+C)

Constraints:
A+C=0
B+2C=0
A+B-C=5
• Nov 25th 2012, 11:39 PM
Prove It
Re: Expand Into Partial Fractions
Quote:

Originally Posted by rrooggeerr
5/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)
multiply by LCD to get:

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

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

5= Ax^2 - A + Bx - B + Cx^2 + 2Cx + C

Group Common Terms:
5= x^2(A+C) + x(B+2C) + (-A-B+C)

Constraints:
A+C=0
B+2C=0
A+B-C=5

This all looks fine. Now solve for A,B,C.
• Nov 25th 2012, 11:44 PM
rrooggeerr
Re: Expand Into Partial Fractions
I get:
A= -5/4
B= -5/2
C= 5/4

I know something is wrong though because it doesn't make the third constraint true :(
• Nov 25th 2012, 11:48 PM
rrooggeerr
Re: Expand Into Partial Fractions
never mind, i got it!!

A = 5/4
B= 5/2
2= -5/4

THANK YOU!
• Nov 26th 2012, 12:05 AM
MarkFL
Re: Expand Into Partial Fractions
There is an alternate method for partial fraction decomposition you may be interested in:

Heaviside cover-up method - Wikipedia, the free encyclopedia