Let $f(x) = x^n-1$ be a polynomial in $\Bbb R[x]$. Factorize $f(x)$ as a product of irreducible polynomials in $\Bbb C[x]$ and show that if $n$ is even, $f(x)$ has two reals roots and if $n$ is odd, $f(x)$ has one real root.

Can someone help me? I know that a polynomium of degree $n$ has exactly $n$ complex roots and that I need to factorize it as a product of $n$ degree $1$ polynomials with $x-a_i$ where $a_i$ is the $i$'th root. But how do I determine the complex roots? And do I have to use the fact that complex roots always come in pairs with its complex conjugated? I know that if $n $is even the are two real roots $1$ and $-1$ and if $n$ is odd the only root is $1$ with multiplicity of $2$.

  • $\begingroup$ This should help : en.wikipedia.org/wiki/Root_of_unity . $\endgroup$
    – LeoDucas
    Nov 25, 2018 at 17:28
  • $\begingroup$ I edited your post to properly $\LaTeX$ify it. Remember to surround your $\LaTeX$ with "\$" signs; thus \$ x^n - 1 \$ yields $x^n - 1$ etc. Cheers! $\endgroup$ Nov 25, 2018 at 17:40

1 Answer 1



The complex roots are the $n$-th roots of unity, which you determine with the exponential form of complex numbers: they have modulus $1$ and their argument must satisfy $$\bigl(\mathrm e^{i\theta}\bigr)^n=\mathrm e^{in\theta}=1=\mathrm e^{i\cdot0}\iff n\theta\equiv 0\mod 2\pi\iff\theta\equiv 0\mod\frac{2\pi}n,$$ taking into account that the complex exponential function has period $2i\pi$.

Can you proceed and show there are $n$ roots of unity and determine which pairs are conjugate?


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