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Formulas and Identities #1 Question 31 Consider the identity \(x^{99}-1 ≡ (x^2-1)P(x) + Cx+D\), where P(x) is a polynomial in x. Find the values of C and D

Mathematics
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C=1, D=-1 ???
Bingo ... How did you get it?!?
silly me... i wrote the proof of it on the margin of my notebook and now i can't find it....:(

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Other answers:

substitute x=1 , and you'll get the equation: 0 = C + D substitute x=-1, and you'll get the equation: -2 = -C + D now solve the system...
..... Silly me....
Well, I don't really see an easy way to do this, so I guess I'll attempt it by "brute force". Let us generically define \(P(x)\) as follows. I think it's pretty clear that it will be of order 97\[P(x) = a_0 + a_1 x + a_2 x^2 + \cdots a_{97} x^{97}\]Multiplying \(x^2-1\) gives us the following.\[(x^2-1)P(x) = x^2P(x)-P(x) =\]\[= -a_0 - a_1 x + (a_0 - a_2)x^2 + \cdots (a_{95}-a_{97})x^{97}-a_{96}x^{98}-a_{97}x^{99} \]Now, let's add \(Cx+D\).\[x^{99}-1=(C-a_0) + (D-a_1) x + (a_0 - a_2)x^2 + \cdots (a_{95}-a_{97})x^{97}-a_{96}x^{98}-a_{97}x^{99}\]By comparing the coefficients on the left and the right, we can write the following system of many equations.\[\begin{align} C-a_0 &= -1 \\D-a_1 &=0 \\ a_0-a_2&=0 \\ &\vdots \\ a_{98}&=0 \\ a_{99}&= 1 \end{align}\]That gives \(C=-1\) and \(D=0\). Hmm.... I'd go with the simpler method above lol.
Thanks!!
Actually, kidding; it does give the same answer...I just was erronous in writing my system of equations... the last two subscripts should be 96 ad 97, which subsequently affect all of the previous equations in a chain which makes the appropriate values of \(C\) and \(D\).
Hmm.. I'll stick to the first method here... since I haven't heard of the second method.. :(
I made it up lmao. It works, though, in theory.

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