# Homework Help: Partial Fraction Decomposition

1. Jul 3, 2010

1. The problem statement, all variables and given/known data
Find the Partial Fraction Decomposition.

$$\frac{4x^2+2x-1}{x^2(x+1)}$$

2. Relevant equations

3. The attempt at a solution

$$\frac{4x^2+2x-1}{x^2(x+1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{(x+1)}$$

$$4x^2+2x-1=x^2(x+1)a+x(x+1)b+x^2(x)c$$

So i can solve for c by plugging in -1 for x but I'm not sure how to solve for a and b.

2. Jul 3, 2010

### vela

Staff Emeritus
You have an extra factor of x on the right side. Fixing that may help you see how to solve for b.

3. Jul 3, 2010

Extra factor? do you mean the right side has too many x? Are you saying that I multiplied by the LCD wrong?

4. Jul 3, 2010

### vela

Staff Emeritus
Yes, when you multiplied by the denominator, you did it incorrectly.

5. Jul 3, 2010

How is that wrong? You need to get an LCD of $$x^3(x+1)$$

So the only way to do that is to multiply a by $$\frac{x^2(x+1)}{x^2(x+1)}$$

b by $$\frac{x(x+1)}{x(x+1)}$$

and c by $$\frac{x^2(x)}{x^2(x)}$$

6. Jul 3, 2010

### Hurkyl

Staff Emeritus
That's not the LCD. Now, it is still a CD so it's fine to multiply by that -- but you need to multiply both sides by the same thing when solving an equation!

P.S. it's just a system of equations.
P.P.S. why not evaluate at 2 or 17 or anything else?

7. Jul 3, 2010

### vela

Staff Emeritus
You may find it easier to think of it as multiplying both sides by the denominator of the LHS:

$$x^2(x+1) \left[\frac{4x^2+2x-1}{x^2(x+1)}\right] = x^2(x+1)\left[\frac{a}{x}+\frac{b}{x^2}+\frac{c}{(x+1)}\right]$$

When you simplify, you'll get

$$4x^2+2x-1 = x(x+1)a + (x+1)b + x^2 c$$

8. Jul 3, 2010

$$4x^2+2x-1=x(x+1)a+(x+1)b+x^2c$$