fraction decomposition..??

GreenTea1mochi · #1 $Old$ November 9th, 2009, 11:52 PM

Us partial fraction decomposition to decompose the given expression: (2x^2+1)/(x^3+2x^2+x)

(2x^2+1)/(x^3+2x^2+x)
= (2x^2+1)/[x(x^2+2x+1)
= (A/x) + (B/(x+1)) + (C/(x+1))
= (2x^2+1)/(x^3+2x^2+x) * (x^3+2x^2+x) = [(A/x) + (B/(x+1)) + (C/(x+1))]*(x^3+2x^2+x)
= 2x^2+1 = A(x^2+1) + B/(x^2+x) + C/(x^2+x)
= 2x^2+1 = A(x^2+1) + (B+C)(x^2+x)

And that's as far as I got. don't know where to go from there..?

Thank you!

CaptainBlack · #2 $Old$ November 10th, 2009, 12:13 AM

Quote:

Originally Posted by GreenTea1mochi $View Post$

(2x^2+1)/(x^3+2x^2+x)

I got up to 2x^2+1 = A(x^2+1) + (B+C)(x^2+x) but don't know where to go from there..?

Thank you!

First you could tell us how you got to where you are, and expressing the question clearly would do no harm either.

I presume that you are required to find the partial fraction decomposition of:

$\frac{2x^2+1}{x^3+2x^2+x}$

First factorise the denominator as much as you can:

$\frac{2x^2+1}{x^3+2x^2+x}=\frac{2x^2+1}{x(x+1)^2}$

Now you need to find $A\ B$ and $C$ such that:

$\frac{2x^2+1}{x^3+2x^2+x}=\frac{2x^2+1}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$

CB

CaptainBlack · #3 $Old$ November 12th, 2009, 01:27 AM

Quote:

Originally Posted by GreenTea1mochi $View Post$

.....help please?

Don't bump.

If you are having difficulties explain what they are, give us a clue.

CB