# The form of a partial fraction decomposition

• Oct 18th 2010, 08:55 AM
rowdy3
The form of a partial fraction decomposition
For the following problem provide the "form" of the partial fraction decomposition for the given fractional expression. You do not have to solve the undetermined coefficients.
4. 2x^2 - 3x + 8 / x^3 + 9x
I took an x out and it's no x(x^2+9) My answer is A/X + BX+C/x^2+9

5. x- 7 / x^4 - 16
I did (x^2+4)(x^2-4). I factor (x^2 - 4) into (x-2)(x+2). My answer is AX+B/x^2+4 + C/(x+2) + D/(x-2)

6. x^2 - 4x + 6 / (x+3)^2 (x^2+1)^2.
My answer is A/(x+3) + B/(x+3) + CX+D/(x^2+1) + EX+F/(x^2+10^2)
Here's a scan of the problems.
http://pic20.picturetrail.com/VOL137.../392720956.jpg
• Oct 18th 2010, 11:33 AM
Soroban
Hello, rowdy3!

Quote:

For the following problems, provide the form of the partial fraction decomposition.

$4.\;\;\dfrac{2x^2 - 3x + 8}{x^3 + 9x}$

My answer: . $\dfrac{A}{x} + \dfrac{Bx+C}{x^2+9}$

$5.\;\;\dfrac{x- 7}{x^4 - 16}$

My answer: . $\dfrac{Ax+B}{x^2+4} + \dfrac{C}{x+2} + \dfrac{D}{x-2}$

$6.\;\;\dfrac{x^2 - 4x + 6}{(x+3)^2(x^2+1)^2. }$

My answer: . $\dfrac{A}{x+3} + \dfrac{B}{(x+3)^2} + \dfrac{Cx+D}{x^2+1} + \dfrac{Ex+F}{(x^2+1)^2}$

Correrct! . . . Good work1
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