Partial Fraction Decomposition

[themadhatter1]

Homework Statement

Find the Partial Fraction Decomposition.

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

Homework Equations

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.

 

[vela]

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

 

[themadhatter1]

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

 

[vela]

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

 

[themadhatter1]

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 $$\frac{x^2(x)}{x^2(x)}$$

 

[Hurkyl]

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

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?

 

[vela]

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

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

 

[themadhatter1]

Haha! your right that isn't the LCD.

ok so now I have:

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

ok so I found c=1 and b=-1 then I can set up a system of equations and find that A=3

Thanks!