Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$.

Question

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.

Answer 1

Without this hint we can see immediately that: $$x^4-2x^3-x^2+2x+1=(x^2-x-1)^2.$$ Can you end it now?

Answer 2

$\left(x-1/x\right)^2=x^2+1/x^2-2$. 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.

Answer 3

$y^2=x^2-2+1/x^2$, so the expression after multiplying by $1/x^2$ can be rewritten $$y^2-2y+1=(y-1)^2=0$$ So $y=1$, or $x-1/x=1$ or $x^2-x-1=0$. Solving this quadratic gives $x=\frac{1\pm\sqrt5}2$, and it is easy to check that $x=0$ doesn't satisfy the original equation, so we have found all solutions.

Answer 4

$Y^2= X^2-2+\dfrac{1}{X^2}$. So $X^2-1+\dfrac{1}{X^2} = Y^2+1$. Then

$$X^2-2X-1+\dfrac{2}{X}+1/{X^2} = Y^2+1-\left(2X-\dfrac{2}{X}\right)=Y^2-2Y+1$$