Let $n \geq 3$ be an integer and $P_n$ the polynomial function such that
$$ P_n\left(x\right) = x^{2n+1}-\left(1+\frac{1}{n}\right)x^{2n}-\left(1+\frac{1}{n}\right)x+1 $$
I need trying to prove that $P_n$ has exactly three real roots.
What I've done is noticing some properties of $P_n$ that enlight me about some real roots it has, however I'm not able to prove there is only three.
First I've noticed : -> $P_n\left(-1\right) = 0$ si $r_1 = -1$ is a real root. -> $P_n\left(0\right) = 1 \geq 0$, $P_n\left(1+\frac{1}{n}\right) = 1-\left(1+\frac{1}{n}\right)^2 \leq 0$ and $\lim\limits_{x \rightarrow +\infty}P_n\left(x\right) = +\infty$ -> The product of all the roots of $P_n$ must be $-1$.
Since $P_n$ is continuous on $\mathbb{R}$, I assume there is one root $r_2$ within $\left[1+\frac{1}{n};+\infty\right[$. Furthermore, the product of all roots (without taking $-1$ into account) must be $1$, so I've guessed that $1/r_2$ must also be a root because complex roots always come by pair.
How can I show that there are no other real roots ? I've tried to obtain the variations of $P_n$ however I was not able to deduce the sign of $P'_n\left(x\right)$. Any help ?
if we diffrentiate we will get $P_n'(x)=(2n+1)\cdot x^{2n}-(2n+2)\cdot x^{2n-1}-(1+\frac {1}{n})$
we will look how many times $P_n'(x)=0$, we can use Alexander Burstein comment to notice we can bound this by 2, as between two roots of the derivite the second derivitive will have a root by rolle's theorem. In the same way we can bound by 3 the number of roots of $P_n$.