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How can I prove that $f(x)= x^6+15x^2-60x+1=0$ can't have three real roots.

First I derive two times for see the signs of derivatives and got that $ 6x^5+30x-60=0$ and $ 30x^4+30=0$ which is always positive but hoe can I conclude that $f$ can't have three roots.

Thanks for your help and time.

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  • $\begingroup$ "Always positive " second derivative means the slope is always increasing. If the polynomial crosses the $y=0$ line three times, what has to happen? $\endgroup$
    – abiessu
    Apr 19 '16 at 18:22
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    $\begingroup$ Also, since it's a polynomial of even degree with real coefficients, it can't have an odd number of real roots. $\endgroup$ Apr 19 '16 at 18:24
  • $\begingroup$ Ok, ok I understand, really thanks $\endgroup$
    – Rachel
    Apr 19 '16 at 18:24
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The second derivate is always positive. Hence, the first derivate can have at most one root. Hence, the function can have at most two roots.

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Hint: Since all coefficients are real, complex roots of the polynomial should go by pairs $x$ and $\bar{x}$.

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  • $\begingroup$ Wow, a very easy way to prove the claim, chapeau! $\endgroup$
    – Peter
    Apr 19 '16 at 18:23
  • $\begingroup$ What if two of roots are real and four imaginary? $\endgroup$ Apr 19 '16 at 18:45
  • $\begingroup$ @RayeesAhmad, according to the problem we only need to show that the polynomial can't have three real roots. $\endgroup$
    – user35603
    Apr 19 '16 at 19:03
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Use rule of change signs The number of positive real roots of $f(x)$ can't exceed the number of change of signs in $f(x)$ and The number of negative real roots of $f(x)$ can't exceed the number of change of signs in $f(-x)$

Now you see $f(x)$ and $f(-x)$ has one and zero sign changes.Hence there can be only one real root but then f(x) will have odd number of imaginary roots which is not possible.Hence $f(x)$ has no real root.

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