Callisto
  • Callisto
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
C=1, D=-1 ???
Callisto
  • Callisto
Bingo ... How did you get it?!?
anonymous
  • anonymous
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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More answers

anonymous
  • anonymous
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...
Callisto
  • Callisto
..... Silly me....
anonymous
  • anonymous
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.
Callisto
  • Callisto
Thanks!!
anonymous
  • anonymous
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 \(D\).
Callisto
  • Callisto
Hmm.. I'll stick to the first method here... since I haven't heard of the second method.. :(
anonymous
  • anonymous
I made it up lmao. It works, though, in theory.

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