Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + 2X + 1 &= 0\\
\Rightarrow X^2 - 2X - 1 + \frac{2}{X} + \frac{1}{X^2} &= 0.
\end{align*}
My problem is that I don't understand how to make the suggested substitution. I'm wondering if there is something obvious I'm missing.
I appreciate any help.

$\left(x-1/x\right)^2=x^2+1/x^2-2$. If you substitute $y=x-1/x$, your equation becomes $y^2+2-2y-1=0$, so $y^2-2y+1=0$, so $y=1$. Then you find $x$ from $x-1/x=1$: $x^2-x-1=0$, $x=(1\pm \sqrt{5})/2$. Each of these is a double root of the original equation.
