I need help finding general formula for factoring these two polynomials: $$P_{2n}(x) = x^{2n} \pm 1$$ $$P_{2n+1}(x) = x^{2n+1} \pm 1$$ where $n$ is an integer. So the goal is to find formula for odd and even polynomials. I tried experimenting with real numbers to find a pattern, but that failed. I only know that complex roots are always in pairs - complex conjugates.

  • $\begingroup$ Do you want to factor these polynomials over $\mathbb C$, over $\mathbb R$, or over an arbitrary ring? $\endgroup$ – Henning Makholm Oct 12 '16 at 20:33
  • $\begingroup$ over C - complex numbers $\endgroup$ – Martin Morris Oct 12 '16 at 20:37


let us do it for $x^{2n}-1$

we have to solve the equation

$z^{2n}=1$ with $z=e^{it}$

so $e^{2int}=e^{2ik\pi}$

which gives the roots

$z_k=e^{i\frac{k\pi}{n}}$ with

$k\in \{0,1,2,...2n-1\}$.



(x^2-2xcos( \frac{ (n-1)\pi }{n})+1)$


Over $\mathbb C$ you don't need to split into odd and even cases -- we always have $$ x^n-1 = (x-1)(x-e^{2\pi i/n})(x-e^{4\pi i/n})\cdots(x-e^{2(n-1)\pi i/n}) $$ and $$ x^n+1 = (x-e^{\pi/n})(x-e^{3\pi i/n})(x-e^{5\pi i/n})\cdots(x-e^{(2n-1)\pi i/n}) $$ because $x^n=1$ and $x^n=-1$ both have $n$ different solutions, so the polynomial is the product of one linear factor $x-\xi$ for each solution $\xi$.

(One of the above factors will always work out to be $x+1$, but whether it is a factor of $x^n+1$ or of $x^n-1$ depends on whether $n$ is odd or even).

  • $\begingroup$ Thank you for your answer, i get it, but I don't know how did you derive the factorization using exponential function. $\endgroup$ – Martin Morris Oct 13 '16 at 12:43
  • $\begingroup$ @MartinMorris You need an understand of what $e^i$ is first, and then it becomes quite obvious. $\endgroup$ – Simply Beautiful Art Oct 24 '16 at 22:23

This is most simply done with some simple algebra:




More specifically, when we are dealing with all complex solutions:

$$x_\phi=e^{2\pi i/\phi}\sqrt[k]a$$

For $\phi\in\{1,2,3,\dots,k\}$. Since you've found all the roots, see the factored form:



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