# fraction decomposition..??

• Nov 9th 2009, 10:52 PM
GreenTea1mochi
fraction decomposition..??
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!
• Nov 9th 2009, 11:13 PM
CaptainBlack
Quote:

Originally Posted by GreenTea1mochi
(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:

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

First factorise the denominator as much as you can:

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

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

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

CB
• Nov 12th 2009, 12:27 AM
CaptainBlack
Quote:

Originally Posted by GreenTea1mochi