Solving Partial Fractions

**transcendentx** · August 23rd 2011, 05:21 AM

Q1)

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

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

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

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

3. $\frac{P(x)}{(ax+b)(cx^2+dx+e)} = \frac{A}{ax+b} + \frac{Bx+C}{cx^2+dx+e}$ where $cx^2+dx+e$ 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.

**Quacky** · August 23rd 2011, 07:08 AM

We have:

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

Taking the discriminant of the quadratic on the denominator:

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

$=16$

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

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

Can you take it from there?

Edit: I'm sorry you had to wait so long for a response.

**transcendentx** · August 23rd 2011, 06:34 PM

Thank youfor your help! No worries about the me taken.

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