# Solving Partial Fractions

• Aug 23rd 2011, 05:21 AM
transcendentx
Solving Partial Fractions
Q1)

$\displaystyle \frac{2x+4}{x^2 + 2x - 3}$

From my understanding, I have to always factorize the denominator of the fraction to one of these 3 formulas to proceed on:

1. $\displaystyle \frac{P(x)}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$

2. $\displaystyle \frac{P(x)}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$

3. $\displaystyle \frac{P(x)}{(ax+b)(cx^2+dx+e)} = \frac{A}{ax+b} + \frac{Bx+C}{cx^2+dx+e}$ where $\displaystyle cx^2+dx+e$ cannot be factored into linear factors (omg i haven't touch on this yet)

However I tried and failed to factorize the denominator to substitute to any of the formulas above.

Please enlighten me! I just need to change them into one of the stated formula and I can to solve it further by myself. Your every effort is appreciated.. Thank you.
• Aug 23rd 2011, 07:08 AM
Quacky
Re: Solving Partial Fractions
We have:

$\displaystyle \frac{2x+4}{x^2+2x-3}$

Taking the discriminant of the quadratic on the denominator:

$\displaystyle 2^2-4(1)(-3)$

$\displaystyle =16$

This tells us that this quadratic will factor, as $\displaystyle 16$ is a perfect square (think of the quadratic formula, and what would happen if you substituted).

Soon, we deduce that the factors are $\displaystyle (x+3)(x-1)$ by trial and error (fortunately, $\displaystyle -3$ doesn't give many options).

Can you take it from there?

Edit: I'm sorry you had to wait so long for a response.
• Aug 23rd 2011, 06:34 PM
transcendentx
Re: Solving Partial Fractions