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Thread: 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 $\displaystyle f$ be any polynomial function. Let:

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

    Where:

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

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

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

    Lets look at one example:


    For

    $\displaystyle f(x) = x^2$

    we know that:

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

    and that:

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

    and also that:

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

    Its obvious that since:

    $\displaystyle (p-q)^2 = p^2 -2pq + q^2$

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

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

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

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

    For some function $\displaystyle g(x)$.

    In other words, prove that $\displaystyle (x-a)^2$ is always a factor of $\displaystyle \Delta_a(x)$ for all polynomial $\displaystyle f$.
    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 $\displaystyle \Delta_a(x) = f(x) - f(a) - (x-a)f'(a)$, it is obvious that $\displaystyle \Delta_a(x)$ is a polynomial. Also, $\displaystyle \Delta_a(a) = 0$, so by the factor theorem $\displaystyle x-a$ is a factor of $\displaystyle \Delta_a(x)$.

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