# Real roots of $x^6+15x^2-60x+1$

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.

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

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