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

Re: Expand Into Partial Fractions

Which way gives you the LCD that is the denominator of the original expression?

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)**

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?

Re: Expand Into Partial Fractions

Option 1 would give me that..right?

Re: Expand Into Partial Fractions

No it wouldn't. Since two of them have as the denominator, the LCD is , NOT as required.

When you have repeated factors, you need one denominator to be and another to be in your partial fraction.

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)]

??

Re: Expand Into Partial Fractions

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

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.

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 :(

Re: Expand Into Partial Fractions

never mind, i got it!!

A = 5/4

B= 5/2

2= -5/4

THANK YOU!

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