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Thread: roots of a 5-degree polynomial

  1. #1
    Senior Member Sambit's Avatar
    Oct 2010

    roots of a 5-degree polynomial

    Here is another equation where I am stuck to find the exact number of real roots: x^5+x^3-2x+1=0. Descarte's rule says it has 0 or 2 +ve real roots and 1 -ve real root. How do I know (apart from Wolframalpha, of course, which says it has no +ve root) how many +ve real roots does this equation have? I see f(0),f(1),f(2) all are >0. But I want something more general to rely on.

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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Feb 2008
    Yuma, AZ, USA
    First note that every odd degree polynomial has at least one real root.

    f(x)=x^5+x^3-2x+1 \implies g'(x)=5x^4+3x^2-2 =(5x^2-2)(x^2+1)

    So the function is increasing on

    \left(-\infty,  \sqrt{\frac{2}{5}}\right) \cup \left(\sqrt{\frac{2}{5}},\infty\right)

    and has a max at


    and a min at


    Note that

    f\left(-\sqrt{\frac{2}{5}} \right) > 0 \text{ and } f\left( \sqrt{\frac{2}{5}}\right) > 0

    So the function has one zero in

    \left(-\infty,  \sqrt{\frac{2}{5}}\right)

    and then is never negative again so it has only one real zero.
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