, 2019

Solution:
Note that neither $0$ nor $1$ are roots of the polynomial. Consider the function
$$Q(x) = \frac{P(x)}{x^{n}} = (2 n) x^{n} + (2 n) x^{-n} + (2 n-1) x^{n-1} + (2 n-1) x^{-n+1} + \cdots + (n+1) x^{1} + (n+1) x^{-1} + n.$$
All $2 n$ of the complex roots of $P(x)$ will be roots of $Q(x)$.
If $|x|=1$, then $x = e^{i \theta}$, and
$$\begin{aligned} Q(x) &= (2 n)\left(x^{n} + x^{-n}\right) + (2 n-1)\left(x^{n-1} + x^{-n+1}\right) + \cdots + (n+1)\left(x + x^{-1}\right) + n \\ &= (2 n)\left(e^{i n \theta} + e^{-i n \theta}\right) + (2 n-1)\left(e^{i(n-1) \theta} + e^{-i(n-1) \theta}\right) + \cdots + (n+1)\left(e^{i \theta} + e^{-i \theta}\right) + n \\ &= (2 n)(2 \cos (n \theta)) + (2 n-1)(2 \cos ((n-1) \theta)) + \cdots + (n+1)(2 \cos (\theta)) + n \end{aligned}$$
which is real. Thus on the unit circle, we have $Q(x)$ is real, and we want to show it has $2 n$ roots there. Rewrite
$$\begin{aligned} P(x) &= (2 n) x^{2 n} + (2 n-1) x^{2 n-1} + \cdots + (n+1) x^{n+1} + n x^{n} + (n+1) x^{n-1} + \cdots + 2 n \\ &= (2 n)\left(x^{2 n} + x^{2 n-1} + \cdots + 1\right) \\ &\quad - \left(x^{2 n-1} + 2 x^{2 n-2} + \cdots + (n-1) x^{n} + n x^{n-1} + (n-1) x^{n-2} + \cdots + 2 x^{2} + x\right) \\ &= 2 n \frac{x^{2 n+1} - 1}{x-1} - x\left(x^{2 n-2} + 2 x^{2 n-3} + \cdots + (n-1) x^{n} + n x^{n-1} + (n-1) x^{n-2} + \cdots + 2 x + 1\right) \\ &= 2 n \frac{x^{2 n+1} - 1}{x-1} - x\left(x^{n-1} + x^{n-2} + \cdots + x + 1\right)^{2} \\ &= 2 n \frac{x^{2 n+1} - 1}{x-1} - x\left(\frac{x^{n} - 1}{x-1}\right)^{2} \end{aligned}$$
and thus
$$Q(x) = \frac{2 n}{x^{n}} \frac{x^{2 n+1} - 1}{x-1} - \frac{x}{x^{n}}\left(\frac{x^{n} - 1}{x-1}\right)^{2}$$
Consider the roots of unity $r_{j} = e^{i \frac{2 \pi}{2 n} j}$, for $j = 0$ to $2 n - 1$. There are $2 n$ such roots of unity: they all have $r_{j}^{2 n} = 1$, and they alternate between those which satisfy $r_{j}^{n} = 1$ or $r_{j}^{n} = -1$. At those $x = r_{j}$, if $r_{j}^{n} = 1$ but $x \neq 1$, then
$$\begin{aligned} Q(x) &= \frac{2 n}{x^{n}} \frac{x^{2 n+1} - 1}{x-1} - \frac{x}{x^{n}}\left(\frac{x^{n} - 1}{x-1}\right)^{2} \\ &= 2 n \frac{x^{1} - 1}{x-1} - x\left(\frac{1-1}{x-1}\right)^{2} = 2 n > 0 \end{aligned}$$
At $x = 1$, we can easily see $Q(1) > 0$.
If $r_{j}^{n} = -1$, then
$$\begin{aligned} Q(x) &= \frac{2 n}{x^{n}} \frac{x^{2 n+1} - 1}{x-1} - \frac{x}{x^{n}}\left(\frac{x^{n} - 1}{x-1}\right)^{2} \\ &= -2 n \frac{x^{1} - 1}{x-1} + x\left(\frac{-1-1}{x-1}\right)^{2} \\ &= -2 n + \frac{4 x}{(x-1)^{2}} \\ &= -2 n + \frac{4}{x - 2 + 1/x} \\ &= -2 n + \frac{4}{2 \cos \left(\frac{2 \pi}{2 n} j\right) - 2} < -2 n - 4 < 0 \end{aligned}$$
since the denominator of this second term is strictly negative $(j \neq 0)$.
Thus at each of the $2 n$-roots of unity, $Q(x)$ alternates in sign, and because $Q(x)$ is real and continuous on the unit circle, it has at least one root between every pair of consecutive roots of unity. Since there are $2 n$ of these pairs, and we know that $Q(x)$ has exactly $2 n$ roots (by the Fundamental Theorem of Algebra), we have found all of $Q$'s roots, and therefore those of $P$.