Partial Fraction Decomposition

[xxwinexx]

Homework Statement

[PLAIN]http://webwork2.math.utah.edu/webwork2_files/tmp/equations/60/0cff8a5107e21ae393dee5038fb6b31.png

Homework Equations

The Attempt at a Solution

I've been attempting to use the general method of solving found at this website:
http://www.purplemath.com/modules/partfrac2.htm

Basically it has me multiply out all of the right side using the LCD. After doing that, I plug in useful numbers for X, such as 0. I must be setting up the LCD for A, B, and C incorrectly. When plugging in 0, I get A=-1, which my homework program is telling me is incorrect.

This is what I got after multiplying through with the LCD:

4x^2-1=A(x+1)(x+1)^2 + B(x)(x+1)^2 + C(x)(x+1)

After multiplying all of those out, and attempting to plug in 0 for x, I get A=-1, which as I've stated is incorrect according to my online course.

Am I setting this up incorrectly? Is my LCD incorrect when multiplying through?

 

[xxwinexx]

Also, shouldn't the RHS of the equation look like this?

A/(2x) + B/(x+1) + C(x+1)^2

As that is all of the terms of the denominator on the LHS?

 

[Mark44]

Also, shouldn't the RHS of the equation look like this?

A/(2x) + B/(x+1) + C(x+1)^2

As that is all of the terms of the denominator on the LHS?

Make that A/(2x) + B/(x+1) + C/(x+1)^2

 

[xxwinexx]

Make that A/(2x) + B/(x+1) + C/(x+1)^2

OK, so now that I know that it was a misprint, should I set up the equation like this?

4x^(2) - 1 = A(x+1)(x+1)^(2) + B(2x)(x+1)^(2) + C(2x)(x+1)

 

[Slats18]

Not quite, the aim of this is to equate denominators across the equality, so you want A,B and C to have the same denominator as the LHS. Then, the denominators can be disregarded for the next part, finding A,B and C.

 

[Mark44]

OK, so now that I know that it was a misprint, should I set up the equation like this?

4x^(2) - 1 = A(x+1)(x+1)^(2) + B(2x)(x+1)^(2) + C(2x)(x+1)

You multiplied both sides of the original equation by 2x(x + 1)². The left side is correct, but the right side isn't. For example, when you multiply A/(2x) by 2x(x + 1)², you should get A(x + 1)², not A(x + 1)³ as you show. In each of the three multiplications, you'll get some cancellation.

Try again, but be more careful.