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Math Help - Proving that (x-a)^2 is always a factor of \Delta(x)?

  1. #1
    Member mfetch22's Avatar
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    Proving that (x-a)^2 is always a factor of \Delta(x)?

    Okay heres the question:



    Let f be any polynomial function. Let:

    \tau_a(x) = f'(a)(x-a)+f(a)

    Where:

    f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

    Also, let the difference between the tangent and f be denoted by \Delta, which is defined:

    \Delta_a(x) = f(x) - \tau_a(x) = f(x) - f'(a)(x-a)-f(a)

    Lets look at one example:


    For

    f(x) = x^2

    we know that:

    \frac{d}{dx} [f(x)] = f'(x) = 2x

    and that:

    \tau_a(x) = (2a)(x-a)+a^2

    and also that:

    \Delta_a(x) = x^2 - (2ax - 2a^2 +a^2) = x^2 - 2ax +a^2

    Its obvious that since:

    (p-q)^2 = p^2 -2pq + q^2

    We know that (x-a)^2 is a factor of \Delta_a(x), specifically that:

    \Delta_a(x) = (1)(x-a)^2

    Now, prove generally, for all polynomial functions f, that \Delta_a(x) can always be written in the form:

    \Delta_a(x) = g(x)(x-a)^2

    For some function g(x).

    In other words, prove that (x-a)^2 is always a factor of \Delta_a(x) for all polynomial f.
    Okay, so I've been trying to do this with polynomial long division and other such attempts, and I just can't figure it out. Any help would be much appreciated. Thanks in advance
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Since \Delta_a(x) = f(x) - f(a) - (x-a)f'(a), it is obvious that \Delta_a(x) is a polynomial. Also, \Delta_a(a) = 0, so by the factor theorem x-a is a factor of \Delta_a(x).

    Now differentiate: \Delta_a'(x) = f'(x) - f'(a). Thus \Delta_a'(a) = 0, and again by the factor theorem x-a is a factor of \Delta_a'(x). But that means that x-a is a repeated factor of \Delta_a(x). In other words, (x-a)^2 divides \Delta_a(x), and so \Delta_a(x) = g(x)(x-a)^2 for some polynomial g(x).
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