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!

## Comments

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TopNewestOr, we can use the 9th roots of unity. – David Lee · 2 years, 4 months ago

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– Daniel Liu · 2 years, 4 months ago

Yes, what happened to the classic roots of unity? Roots are just \(e^{2ki\pi/9}\)with \(k=1\to 8\).Log in to reply

– Nathan Ramesh · 2 years, 4 months ago

Except all roots of unity give you is \(e^{\frac{2\pi i}{9}}\) and not a numerical answer with radicals. The method here allows you to compute \(\sin {40^{\circ}}\)Log in to reply

– David Lee · 2 years, 4 months ago

Or you can also just do some Euler's formula with \( e^{\dfrac{2\pi i}{9}} \).Log in to reply

– Nathan Ramesh · 2 years, 4 months ago

ummmm which euler's formula?Log in to reply

– Curtis Clement · 1 year, 9 months ago

\(e^{i\phi}\) = cos\(\phi\) + \({i}\)sin\(\phi\) ... substituting \(\pi\) gives \(e^{i\pi}\) = -1, where \({i}\) = \(\sqrt{-1}\)Log in to reply

– Aman Sharma · 1 year, 9 months ago

Where i can find this bookLog in to reply

– Daniel Liu · 2 years, 4 months ago

\(e^{i\theta}=\text{cis}(\theta)\)Log in to reply

– Nathan Ramesh · 2 years, 4 months ago

How does that let you solve for \(\sin{40^{\circ}}\)Log in to reply

wow – Daniel Liu · 2 years, 4 months ago

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– Daniel Liu · 2 years, 4 months ago

:oLog in to reply

– David Lee · 2 years, 4 months ago

2 DanielsLog in to reply

– Sharky Kesa · 2 years, 4 months ago

Oh God, this is not good. One Daniel is enough for me, but two? Double Trouble.Log in to reply

– Jubayer Nirjhor · 2 years, 4 months ago

2 LiusLog in to reply

– Ahaan Rungta · 2 years, 4 months ago

2 14-year-old-Daniel-Lius. Enough to get a conversation off-topic.Log in to reply

– Jubayer Nirjhor · 2 years, 4 months ago

2 14-year-old-Daniel-Lius-residing-in-USA. conversation closed.Log in to reply

Where are all other pages? – Megh Choksi · 1 year, 9 months ago

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Yes I used the same to expand cosnx – Megh Choksi · 1 year, 9 months ago

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@Nathan Ramesh Can you add this to the Roots of Unity Applications Wiki page?

Select "Write a summary", and then copy-paste your text into it (with minor formatting adjustments if relevant. Thanks! – Calvin Lin Staff · 2 years ago

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– Nathan Ramesh · 2 years ago

Done! I put it at the bottom. Let me know if it is bugged (I posted it from my phone). Thanks!Log in to reply

I reshared your note. :D – Finn Hulse · 2 years, 4 months ago

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