Number of roots of the equation $x^{2016}-x^{2011}+x^{2006}-x^{2001}+1=0$

Question

First,i apply rational root theorem as the leading and trailing coefficient is 1. Therefore possible roots are +1 or -1. None of them satisfy.

Then i use descrates rule which tells me about at most four positive real roots and no negative real root.

Can someone explain me to find the exact number of roots.

Answer 1

Hint: You have already shown that there are no zeros for $x < 0$.

For $x > 1$, we have $x^{2016} > x^{2011}$, $x^{2006} > x^{2001}$, and $1 > 0$.

For $0 < x < 1$, we have $x^{2016} > 0$, $x^{2006} > x^{2011}$, and $1 > x^{2001}$.

Can you use these facts to show that there are no real roots?