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Thread: roots of a 5-degree polynomial

  1. #1
    Senior Member Sambit's Avatar
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    roots of a 5-degree polynomial

    Here is another equation where I am stuck to find the exact number of real roots: $\displaystyle x^5+x^3-2x+1=0$. Descarte's rule says it has $\displaystyle 0$ or $\displaystyle 2$ +ve real roots and $\displaystyle 1$ -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 $\displaystyle f(0),f(1),f(2)$ all are $\displaystyle >0$. But I want something more general to rely on.

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

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

    So the function is increasing on

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

    and has a max at

    $\displaystyle x=-\sqrt{\frac{2}{5}}$

    and a min at

    $\displaystyle x=\sqrt{\frac{2}{5}}$

    Note that

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

    So the function has one zero in

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

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