Partial Fractions . . . HELP!!!

**potato92** · April 15th 2010, 06:23 PM

Find the partial fractions for the following:

3x² - 2x + 8
------------------------
x³ + 2x² + 4x + 8

9x² + 21x - 24
--------------------------
x³ + 4x² - 11x - 30

My biggest problem with these is that I have trouble factoring the bottom cubic in the second problem, but I'm having trouble with everything else also. Every time I try to solve the problems I end up with a different answer than the time before. I don't know what I am doing wrong!

I shall be eternally grateful to whoever helps me solve these problems. So grateful, in fact, that I shall bestow upon them this bear hug.

. . . Wow. It's obviously sleepy time for me. My true weirdness is oozing into your website. Terribly sorry about that.

**harish21** · April 15th 2010, 06:34 PM

Originally Posted by potato92

Find the partial fractions for the following:

3x² - 2x + 8
------------------------
x³ + 2x² + 4x + 8

9x² + 21x - 24
--------------------------
x³ + 4x² - 11x - 30

My biggest problem with these is that I have trouble factoring the bottom cubic in the second problem, but I'm having trouble with everything else also. Every time I try to solve the problems I end up with a different answer than the time before. I don't know what I am doing wrong!

I shall be eternally grateful to whoever helps me solve these problems. So grateful, in fact, that I shall bestow upon them this bear hug.

. . . Wow. It's obviously sleepy time for me. My true weirdness is oozing into your website. Terribly sorry about that.

Question 2:

Hint:

$\frac{9x^2+21x-24}{x^3+4x^2-11x-30} = \frac{{9x^2+21x-24}}{(x+2)(x+5)(x-3)}$

You need to find the partial fraction for the above term:

Try:

$\frac{{9x^2+21x-24}}{(x+2)(x+5)(x-3)} = \frac{A}{x+2} + \frac{B}{x+5}+\frac{C}{x-3}$

solve this now. You should deninitely have partial fractions in your book/notes. question 1 would also be similar to this. if you cannot solve, show your work on where you get stuck!

**potato92** · April 15th 2010, 06:55 PM

Thanks harish! I think those factors were what I needed.

**wonderboy1953** · April 16th 2010, 10:42 AM

If your problem had a 1 in the numerator, there's a nice formula to bypass messy algebra which I'll demonstrate in this specific case:

$[1/(x + 2)(x + 5)(x - 3)]$ = $A/(x + 2) + B/(x + 5) + C/(x - 3)$

$A = 1/(5 - 2)\cdot(-3 - 2)$
$B = 1/(2 - 5)\cdot(-3 - 5)$
$C = 1/(2 - -3)\cdot(5 - -3)$

This formula works whenever you have a 1 in the numerator and linear factors in the denominator.

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