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Thread: p(x) is divisible by (ax+b)^n

  1. #1
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    p(x) is divisible by (ax+b)^n

    Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
    A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
    Now my p(x) = (ax+b)^n-1 s(x)
    then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
    p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
    p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
    but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?

  2. #2
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    Quote Originally Posted by dan34 View Post
    Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
    A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
    Now my p(x) = (ax+b)^n-1 s(x)
    then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
    p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
    p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
    but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?
    [tex](3x + 2)^2 \ | \ p(x) = (3x + 2)^2 \text { and } p'(x) = 6(3x + 2).[/tex]

    But it is not generally true that [tex](3x + 2)^{(100-1)} \ | \ 6(3x + 2).[/tex]

    Tough to prove something that is false!

  3. #3
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by dan34 View Post
    Show that if the pol. p(x) is divisible by (ax+b)^2, then p'(x) is divisible by (ax + b)^(n-1) where n > 1
    A polynomial f(x) to be divisible by another polynomial g(x), there should be another polynomial s(x), s.t f(x) = g(x)s(x)
    Now my p(x) = (ax+b)^n-1 s(x)
    then my p'(x) = ((ax+b)^n)' s(x) + ((ax+b)^n) (s(x))'
    p'(x) = n(ax+b)^(n-1) * (ax + b)' s(x) + ((ax+b)^n) (s(x))
    p'(x) = n(ax+b)^(n-1) * a * s(x) + ((ax+b)^n) (s(x))
    but I do not know how to proceed anymore? Can anybody give me a hint? And about the s(x), should I leave the s(x) just like this or I have to give an expression (value) to it?
    Your subject line refers to "(ax + b)n", rather than to "(ax + b)2". Is the square in the "show" statement a typo for a power of "n"?

    Thank you!

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