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Exercise: Write $x^5-1$ as product of real polynomials of rate (max) 2.

I would start as $(x-1)(x^4+ x^3+x^2+x+1)$, but then I have no idea how to continue. Till now I had only even terms in parenthesis and I managed. Now I cannot get the solution. Please help, because I really have no idea how to continue. Is there maybe an algorithm for solving this?

José Carlos Santos
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continuity
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2 Answers2

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dividing by $x^2\ne 0$ we get $$x^2+\frac{1}{x^2}+x+\frac{1}{x}+1$$ now Substitute $$x+\frac{1}{x}=t$$

Dr. Sonnhard Graubner
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The roots of $x^5-1$ are $1$, $e^{\pm\frac{2\pi i}5}$, and $e^{\pm\frac{4\pi i}5}$. Therefore,\begin{align}x^5-1&=(x-1)\left(x-e^{\frac{2\pi i}5}\right)\left(x-e^{\frac{-2\pi i}5}\right)\left(x-e^{\frac{4\pi i}5}\right)\left(x-e^{\frac{-4\pi i}5}\right)\\&=(x-1)\left(x^2-2\cos\left(\frac{2\pi}5\right)x+1\right)\left(x^2-2\cos\left(\frac{4\pi}5\right)x+1\right).\end{align}

Trevor Gunn
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José Carlos Santos
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  • Your final result is a product of 3 degree-1 polynomials. Are you sure you did the math right? Perhaps you dropped a square. – BRSTCohomology Jan 01 '18 at 18:20
  • Looks like it's been fixed. – BRSTCohomology Jan 01 '18 at 18:22
  • Can you please explain, how did you get the roots with e in it? – continuity Jan 01 '18 at 18:23
  • @continuity It is clear that $\left(\exp\left(\frac{2\pi ki}5\right)\right)^5=\exp(2\pi ki)=1$. On the other hand, if you replace $k$ with $0$, $1$, $2$, $3$, or $4$, you'll get $5$ distinct fifth roots of $1$, and therefore there can be no more. Finally, $\exp\left(\frac{8\pi i}5\right)=\exp\left(-\frac{2\pi i}5\right)$, and $\exp\left(\frac{6\pi i}5\right)=\exp\left(-\frac{4\pi i}5\right)$. – José Carlos Santos Jan 01 '18 at 18:30
  • Didn't even realize. Thank you! – continuity Jan 01 '18 at 18:32