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I told you.
Looks right to me! Let us know if anything about the technique you used doesn't make sense.
I went back to this question and tried it out myself.
I cannot divide 3x^4 + 7x^3 + x^2 - 7x - 4 by x^2 - 1 directly (not using my method of long division at least). If I divide by (x + 1) I am left with a cubic expression 3x^3 + 4x^2 - 3x - 4. Do I need to divide this by the other factor (x - 1)?
Using the difference of 2 squares rule, (x² - 1) = (x+1)(x-1).
how do I divide f(x) by x^2-1?
Good. but if a=7 and b=1 we get:
long division of x^2 - 1/f(x) just like you said krassi?
Nope, no idea
(x^2 - 1) / (x - 1) = x + 1!
I do indeed know how to find a and b if I have two factors but I really have no idea how to get the second factor. What must I do to x - 1 to make x^2 - 1?
One of the factors of x^2-1 is what you were using before: x-1. You should be able to figure out the other one either by division or by trial and error, the answer isn't complicated.
How do I do this? I've never seen such a factor before.
Ah, that would explain it then. Saying something has (x^2 - 1) as a factor is actually saying it has two different factors - that is, anything that is a factor of (x^2-1) must also be a factor of the equation you're trying to solve for. Try breaking (x^2-1) up into its constituent factors and applying them part by part and see if that gets you anywhere. Good luck, and feel free to ask for more help if you get stuck with that.