Originally Posted by

**sidhlyn** Gosh....thank you so much....

But....

"This leads to the system

a+b+c=0

c-b=0

-4a=8 "

is the very part I don't get.

I don't know how you got these figures. ?

I don't understand how to get to (A+B+C)y^2 + 2(C-B)y-4A?

Where did the extra A, B, & C come from.

I'm sorry....our instructer did not show us this way, but the textbook has it set up like how you've explained.

The textbook does not break down HOW to do that either.

They just print it as you did & then gives the "systems." How do you get to the systems?

I did get the antiderivative finally, but I really WANT to know HOW to do those necessary middle steps that I'm missing.

LOL- I think I was going about it the "long" way and was able to come up with A= -4 & C=1, but got stumped on solving for B because there's no way to zero out the C with "y^2= -2y" from: 8=A(y^2-4)=B(y^2-2y)+C(y^2+2y) with plugging in "y" values as we were shown by our instructor.

When you get the chance, can I please have the breakdown on how to get those "systems?" <Anyone out there who has the patience to list that out- please?> I appreciate your help & can use the info provided on my homework, but I have just got to know how to do that. I'm sure it's probably something simple I'm not seeing.

Thanks a million!!