Help!! Im a little stuck.. I need to decompose this into partial fractions:

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Help!! Im a little stuck.. I need to decompose this into partial fractions:

Mathematics
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\[(2x^2-11x+5)/(x-3)(x^2+x-4)\]
What course are you currently taking?
This is Pre-Calc Hero

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Okay, just wondering.
Then I use A/(x-3)+B(x^2+x-4)
What textbook are you using for the course?
Precalculus: Graphs and Models by Addison Wesley
Thanks
Its pretty old. 1997
That's okay. Sometimes, old is better.
Okay. I mean its probably all the same techniques, for the most part.
What edition?
Well, nevermind. I'll get get the latest edition.
Ok I wasnt sure. I dont know if it says it for sure.
If you have the book, it will say what edition it is. You just have to look, but okay.
I have about 9 min left before it finishes downloading
I really dont see it at all :O This is the exact book though http://www.abebooks.com/book-search/isbn/0201694425/hard-cover/
Ok thats fine. Thanks
If you want to show someone a book, use amazon.com
ahh alright.
Yes, that version is very old. It's not even available for download. It's the first edition apparently
4 minutes left..
ok
If you have the book, is there any reason why you are not able to understand this concept?
I get the concepts down sort of, I just end up with the wrong numbers. I either do a step too early or Im not doing it completely right. I just want to make sure I understand it
Okay, I will have a look here in about 15 secs
Ok Ill go through with what Im doing and you can maybe see what Im doing wrong
Let me know when its done
It might take me a bit to figure this out.
Ok thats fine!
Okay, I see something similar to this where it talks about setting up a decomposition template.
Its in a lesson called Partial Fractions in my book
That sounds right though
\[\frac{2x^2 - 11x+5}{(x-3)(x^2+x-4)} = \frac{A}{x-3} + \frac{Bx + C}{x^2 + x -4}\]
Maybe that's what you're looking for.
So then its \[A(x^2+x-4)+(x-3)(Bx+C)=2x^2-11x+5\]
What I posted is in your book so it is correct. If what you posted is the next step, then by all means, go right ahead, but maybe someone else can take a look at that for you.
Ok thats cool. I needed that for sure so thanks a lot Hero

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