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 $2$.