# Partial Fraction Decomposition

• Mar 14th 2013, 03:23 PM
Steelers72
Partial Fraction Decomposition
All I need to do is write the form of the partial function (I don't have to find the numerical values) of this:

x2/(x2-2x-2) partial fraction decomposition

I know the rule of quadratic factor is Ax+B/ax2+bx+c

so wouldn't it be Ax+B/x2+2x+2 or am I missing something? Kind of confused...
• Mar 14th 2013, 04:50 PM
Soroban
Re: Partial Fraction Decomposition
Hello, Steelers72!

I don't understand the question.
The problem can be solved without Partial Fractions.

Quote:

All I need to do is write the form of the partial function.
I don't have to find the numerical values.

. . . $\displaystyle \frac{x^2}{x^2 -2x-2}$

First of all, the fraction is "improper" . . . Use long division.

. . $\displaystyle \frac{x^2}{x^2-2x-2} \;=\;1 + \frac{2x + 2}{x^2-2x-2}$

Second, are we suppose to factor that quadratic?
. . . . $\displaystyle x^2 - 2x - 2 \;=\;\big[x - (1+\sqrt{3})\big]\,\bog[x - (1-\sqrt{3})\big]$

. . Then work on: .$\displaystyle \frac{2x+2}{x^2-2x-2} \;=\;\frac{A}{x-1-\sqrt{3}} + \frac{B}{x - 1 + \sqrt{3}}$ .(ugh!)

Or should we solve: .$\displaystyle \frac{2x+2}{x^2-2x-2} \;=\;\frac{Ax}{x^2-2x-2} + \frac{B}{x^2-2x-2}$

Would it surprise you to learn that $\displaystyle A = 2$ and $\displaystyle B = 2\,?$
• Mar 14th 2013, 09:33 PM
Steelers72
Re: Partial Fraction Decomposition
Wow! Thank you for that brilliant walk-through! Sorry for the question wording; it was asking just to find out the form of the partial fraction decomposition of the function

ie. Given 5x/x^2 +x -6

= A/x+3 B/x-2

That's all they wanted