If$ f(x)=x^{10000}-x^{5000}+x^{1000}+x^{100}+x^{50}+x^{10}+1$, what is the number of rational roots of $f(x)=0$?

Question

The question is:

If$$ f(x)=x^{10000}-x^{5000}+x^{1000}+x^{100}+x^{50}+x^{10}+1$$ what is the number of rational roots of $f(x)=0$?

I used descrates rule.

As number of time sign change is two therefore positive real roots is less than $2$ .

Also the function is even number of negative real roots is also less than $2$. But it gives me information about real roots not rational.

Answer 1

By the rational root theorem, the rational roots of this polynomial must be integers, because the polynomial is monic. Moreover, the integer roots must divide $1$ and so must be $1$ or $-1$. Neither is actually a root.

Answer 2

If $\frac pq$ is a root (in simplest form), where $q \neq 1$, then observe that $f(x)$ cannot be zero (or even integer) because $x^{10000}$ will have a denominator of $q^{10000}$.

It remains to check integers; you can easily check $-1, 0, 1$ yourself, and $|x| \geq 2$ fails since $x^{10000}$ will be much larger than the rest of the terms.