# Irreducible fatorization of $X^n-1$ in both the ring of polynomials with complex coefficients and real coefficients

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 $$2$$.

• This should help : en.wikipedia.org/wiki/Root_of_unity . Nov 25 '18 at 17:28
• 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! Nov 25 '18 at 17:40

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?