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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!

Note by Nathan Ramesh
3 years ago

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Or, we can use the 9th roots of unity. David Lee · 3 years ago

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@David Lee Yes, what happened to the classic roots of unity? Roots are just \(e^{2ki\pi/9}\)with \(k=1\to 8\). Daniel Liu · 3 years ago

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@Daniel Liu 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 \(\sin {40^{\circ}}\) Nathan Ramesh · 3 years ago

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@Nathan Ramesh Or you can also just do some Euler's formula with \( e^{\dfrac{2\pi i}{9}} \). David Lee · 3 years ago

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@David Lee ummmm which euler's formula? Nathan Ramesh · 3 years ago

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@Nathan Ramesh \(e^{i\phi}\) = cos\(\phi\) + \({i}\)sin\(\phi\) ... substituting \(\pi\) gives \(e^{i\pi}\) = -1, where \({i}\) = \(\sqrt{-1}\) Curtis Clement · 2 years, 6 months ago

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@Nathan Ramesh Where i can find this book Aman Sharma · 2 years, 6 months ago

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@Nathan Ramesh \(e^{i\theta}=\text{cis}(\theta)\) Daniel Liu · 3 years ago

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@Daniel Liu How does that let you solve for \(\sin{40^{\circ}}\) Nathan Ramesh · 3 years ago

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wow Daniel Liu · 3 years ago

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@Daniel Liu :o Daniel Liu · 3 years ago

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@Daniel Liu 2 Daniels David Lee · 3 years ago

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@David Lee Oh God, this is not good. One Daniel is enough for me, but two? Double Trouble. Sharky Kesa · 3 years ago

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@David Lee 2 Lius Jubayer Nirjhor · 3 years ago

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@Jubayer Nirjhor 2 14-year-old-Daniel-Lius. Enough to get a conversation off-topic. Ahaan Rungta · 3 years ago

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@Ahaan Rungta 2 14-year-old-Daniel-Lius-residing-in-USA. conversation closed. Jubayer Nirjhor · 3 years ago

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Where are all other pages? Megh Choksi · 2 years, 6 months ago

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Yes I used the same to expand cosnx Megh Choksi · 2 years, 6 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, 9 months ago

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@Calvin Lin Done! I put it at the bottom. Let me know if it is bugged (I posted it from my phone). Thanks! Nathan Ramesh · 2 years, 9 months ago

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I reshared your note. :D Finn Hulse · 3 years ago

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