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Math Help - Polynomial question 2

  1. #1
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    Polynomial question 2

    The remainder obtained when the polynomial x+x^3 +x^9+x^27+x^81+x^2^243 is divided by x^2-1 is:

    The answer is given as 6x but I dont know how to proceed.

    Thanks
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  2. #2
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    Poland
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    Shouldn't it be: P(x)=x+x^3 +x^9+x^{27}+x^{81}+x^{243}?

    Anyway, when divided by x^2-1=(x-1)(x+1):

    P(x)=[Q(x)](x-1)(x+1) + R(x), where Q(x) is the quotient, R(x) is the remainder. Its degree can be at most 1, so R(x)=ax+b and

    P(x)=[Q(x)](x-1)(x+1) + ax+b
    Now
    P(1)=1+1^3+1^9+1^{27}+1^{81}+1^{243}=6
    P(-1)=-1+(-1)^3+(-1)^9+(-1)^{27}+(-1)^{81}+(-1)^{243}=-6

    but also P(1)=a+b
    P(-1)=-a+b

    Now you've got 2 simultaneous equations, solving for a and b gives a=6, b=0 so R(x)=6x.
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  3. #3
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    nopes . x^2^243 only

    anyways, will try with your method and c if i can get an answer.
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