# partial fractions

• Mar 11th 2010, 09:03 AM
satis
partial fractions
can someone point me in the right direction on this?

$\displaystyle \frac{8x^3+10x^2 + 54x + 72}{x^4 - 81}$

I factored the denominator to

$\displaystyle \frac{8x^2+10x^2+54x+73}{(x^2+9)(x^2-9)}$

so breaking it up I get

$\displaystyle \frac{Ax+B}{x^2+9}+ \frac{Cx+D}{x^2-9}$

which turns into

$\displaystyle 8x^3 + 10x^2 + 54x + 72 = (Ax+B)(x^2+9) + (Cx+D)(x^2+9)$

which 'simplifies' to

$\displaystyle 8x^3 + 10x^2 + 54x + 72 = Ax^3 + Cx^3 + Bx^2 + Dx^2 - 9Ax + 9Cx - 9B + 9D$

which translates into the following system of equations

$\displaystyle 8= A + C$
$\displaystyle 10 = B + D$
$\displaystyle 54 = -9A + 9C$
$\displaystyle 72 = -9B + 9D$

Assuming I did all this right, I'm not sure where to go. I'm used to at least one of those systems to actually be solvable... then you can run through the rest with substitution. In this case, I either made a mistake, or I need a bit of guidance on how to proceed.

*edit*
ok, so it occurred to me I could partially solve one of the above and then sub... so I solved 8 = A + C to be A = C- 8. I then subbed that into the third equation in the list... but end up with 54 = 72. So I'm thinking I must have screwed something up somewhere in there.

*edit again*
ok, so my math sucks. I was able to solve for C... I'm going to go through the rest now and see if I can figure it out. Pardon if I wasted anyone's time. I'll post back either way.
• Mar 11th 2010, 10:02 AM
A+c=8
9(c-a)=54
c-a=6
c+a=8
2c=14, c=7, a=1

b+d=10
9(d-b)=72
d-b=8
d+b=10
2d=18, d=9, b=1
• Mar 11th 2010, 12:38 PM
Opalg
Quote:

Originally Posted by satis
can someone point me in the right direction on this?

$\displaystyle \frac{8x^3+10x^2 + 54x + 72}{x^4 - 81}$

I factored the denominator to

$\displaystyle \frac{8x^2+10x^2+54x+73}{(x^2+9)(x^2-9)}$

so breaking it up I get

$\displaystyle \frac{Ax+B}{x^2+9}+ \frac{Cx+D}{x^2-9}$

It's probably best to factorise $\displaystyle x^2-9$ as $\displaystyle (x+3)(x-3)$, so that your partial fraction becomes

$\displaystyle \frac{Ax+B}{x^2+9}+ \frac{C}{x+3} + \frac{D}{x-3}$.

You may find it easier to find the coefficients in that form.
• Mar 11th 2010, 01:52 PM
Soroban
Hello, satis!

Quote:

$\displaystyle F \;=\;\frac{8x^3+10x^2 + 54x + 72}{x^4 - 81}$

I factored the denominator to: .$\displaystyle \frac{8x^3+10x^2+54x+73}{(x^2+9)(x^2-9)} \begin{array}{c} \\ _{\Leftarrow\:\text{ This factors further!}} \end{array}$

The denominator factors to: .$\displaystyle (x-3)(x+3)(x^2+9)$

Your set-up should have been: .$\displaystyle \frac{8x^2+10x^2+54x + 72}{(x-3)(x+3)(x^2+9)} \;=\;\frac{A}{x-3} + \frac{B}{x+3} + \frac{Cx+D}{x^2+9}$

. . which has solutions: .$\displaystyle A = 5,\;B = 2,\;C = 1,\;D = 1$

Therefore: .$\displaystyle F \;=\;\frac{5}{x-3} + \frac{2}{x+3} + \frac{x+1}{x^2+9}$

Opalg beat me to it . . . *sigh*
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