Partial Fraction Decomposition

[themadhatter1]

Homework Statement

Find the Partial Fraction Decomposition.

[tex]\frac{4x^2+2x-1}{x^2(x+1)}[/tex]

Homework Equations

The Attempt at a Solution

[tex]\frac{4x^2+2x-1}{x^2(x+1)}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{(x+1)}[/tex]

[tex]4x^2+2x-1=x^2(x+1)a+x(x+1)b+x^2(x)c[/tex]

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 [tex]x^3(x+1)[/tex]

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

b by [tex]\frac{x(x+1)}{x(x+1)}[/tex]

and c by [tex]\frac{x^2(x)}{x^2(x)}[/tex]

 

[Hurkyl]

How is that wrong? You need to get an LCD of [tex]x^3(x+1)[/tex]

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:

[tex]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][/tex]

When you simplify, you'll get

[tex]4x^2+2x-1 = x(x+1)a + (x+1)b + x^2 c[/tex]

 

[themadhatter1]

Haha! your right that isn't the LCD.

ok so now I have:

[tex]4x^2+2x-1=x(x+1)a+(x+1)b+x^2c[/tex]

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

Thanks!