Polynomials sprint: How to solve a seemingly disgusting polynomial

On page 248, it is shown how to solve the polynomial $x^4+x^3+x^2+x+1=0$ . In this note, I will explain how to solve the polynomial $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ First we multiply both sides by $(x-1)$ to get $x^9-1=0\implies x^9=1$ Then since $x^9=1$ , we can divide some terms by $x^9$ $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=$ $\dfrac{x^8}{x^9}+\dfrac{x^7}{x^9}+\dfrac{x^6}{x^9}+\dfrac{x^5}{x^9}+x^4+x^3+x^2+x+1=$ $\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+\dfrac{1}{x^4}+x^4+x^3+x^2+x+1$ $\left(x+\dfrac{1}{x}\right)+\left(x^2+\dfrac{1}{x^2}\right)+\left(x^3+\dfrac{1}{x^3}\right)+\left(x^4+\dfrac{1}{x^4}\right)+1=0$ Now we let $y=x+\dfrac{1}{x}$ so we have $(y)+(y^2-2)+(y^3-3y)+((y^2-2)^2-2)+1=0$ which after simplification becomes $y^4+y^3-3y^2-2y+1=0$ Using the Rational Root Theorem (discussed on page 246) we quickly find that $y=-1$ is a root and factor the polynomial as $(y+1)(y^3-3y+1)$ Now the rest is simple! We can use the cubic solving method discussed on pages 247-248 to find the roots of $y^3-3y+1$ Then simple quadratic bashing gives us our roots!

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Polynomials sprint: How to solve a seemingly disgusting polynomial

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