# Thread: another fraction to decompose

1. ## another fraction to decompose

(x^2+2x+8) / (x^3-4x^2+4x)

2. Originally Posted by cityismine
Express $\frac{x^2+2x+8}{x^3-4x^2+4x}$ as a sum of partial fractions

first factor the denominator $\frac{x^2+2x+8}{x(x-2)^2}$

$\frac{x^2+2x+8}{x(x-2)^2} \equiv \frac{A}{x} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$

Now continue doing this with the method your familiar with.

Bobak

3. Originally Posted by cityismine
(x^2+2x+8) / (x^3-4x^2+4x)
Factor the Denominator first:

$x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2$

Thus, the PFD is :

$\frac{x^2+2x+8}{x(x-2)^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$

Try taking it from here.

4. 2/x - 1/(x-2) + 8/(x-2)2

Is that the right answer?

5. Originally Posted by cityismine
2/x - 1/(x-2) + 8/(x-2)2

Is that the right answer?
Yes, Good job

Bobak

6. OMG, the book was showing a different answer, so I thought it was wrong. I guess the book is wrong.