# Thread: More Partial Fractions Help

1. ## More Partial Fractions Help

Well, I took a quiz earlier this week, and I didn't do so hot on it (59). One of the questions I missed was a partial fraction question. Under normal circumstances I wouldn't be on here asking about it, but we have a problem similar to the one seen on the quiz (which I will show you in a moment) on the practice test, and there's a chance it could be on the test this coming Friday.

$\displaystyle \frac{1}{4x^2-9}$

~~~~~~~

Obviously this fraction is improper, so I divided it out to get

$\displaystyle 1+\frac{4x^2+0x-9}{4x^2+0x-9}$.

I got stuck after this point. When I reviewed it after I turned in the quiz. $\displaystyle \frac{4x^2+0x-9}{4x^2+0x-9}$ cancels out, leaving 1 + 1, which equals two. I'm thinking this is wrong, but this is how my thinking went about it, at least on the quiz.

Just now when I looked at it again, I realized it was actually $\displaystyle 1 + \frac{-9}{4x^2+0x-9}$. But I'm still unsure as to how to go about doing it from there.

I appreciate any and all help given to this problem.

2. Note that $\displaystyle (2x+3)-(2x-3)=6.$ (Pretty obvious but useful.) Now why did I do this? 'cause you can replace it in the numerator and split the original ratio into two fractions.

3. I see how you were able to factor that expression, but I have absolutely no idea where you got the six from.

EDIT: Never mind, just figured it out.