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Math Help - Harder Polynomial Question

  1. #1
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    Exclamation Harder Polynomial Question

    Show that (x-1)(x-3) is a factor of P(x) = x^p(3^q-1)+x^q(1-3^p)+(3^p-3^q) where p and q are positive integers.
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Show that (x-1)(x-3) is a factor of P(x) = x^p(3^q-1)+x^q(1-3^p)+(3^p-3^q) where p and q are positive integers.
    Show us your work so far and we can help you where you go wrong.

    Have you tried subbing in the values of 1 and 3, see what you get?
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  3. #3
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    Exclamation

    Ummm

    P(1) = 1^p.3^q - 1^p + 1^q -1^q.3^p+3^p-3^q
    P(3)=0

    I've got these but I have no idea how to use that to demonstrate what is required ...

    Is it because P(3) = 0, it means (x-3) is a factor and SO (x-1)(x-3) is???
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  4. #4
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    Hello, xwrathbringerx!

    You had it . . . and didn't know it.


    Show that (x-1)(x-3) is a factor of: . P(x) =\: \:x^p(3^q-1) + x^q(1-3^p) + 3^p-3^q
    where p and q are positive integers.
    Fact: .If P(a) = 0, then (x-a) is a factor of P(x).


    P(1) \:=\:1^p(3^q-1) + 1^q(1-3^p) + 3^p - 3^q \;=\; 3^q - 1 + 1 - 3^p + 3^p - 3^q \;=\;0

    . . Therefore, (x-1) is a factor of P(x).


    P(3) \;=\;3^p(3^q - 1) + 3^q(1-3^p) + 3^p - 3^q \;=\;3^{p+q} - 3^p + 3^q - 3^{p+q} + 3^p - 3^q \;=\; 0

    . . Therefore, (x-3) is a factor of P(x).

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