Factoring $p(x) = x^n -1$ for any natural number $n$

Question

Can I say that from inspection, $(x-1)$ is a factor which implies that $p(1) = 0$.

I then used long division which then gave this: $p(x) = (x-1)(\sum \limits _{i=1} ^{n} x^{n-i})$

How would you have approached the (non) trivial problem?

Answer 1

I would recommend reading up on cyclotomic polynomials. These are the minimal polynomials of primitive roots of unity.

Using the theory of cyclotomic polynomials, this problem is straightforward.

$$x^n-1 = \prod_{d|n} \Phi_d(x) $$

where $\Phi_d(x)$ is the $d^{th}$ cyclotomic polynomial.

The Wikipedia article lists a bunch of the cyclotomic polynomials, and there is also an easy formula for computing them.

Answer 2

This is a high school identity, with a very easy proof by induction:

Suppose $x^n-1=(x-1)(x^{n-1}+\dots+x+1)$ for some $n$. Then \begin{align*}x^{n+1}-1&=(x^{n+1}-x^n)+(x^n-1)\\&=x^n(x-1)+(x-1)(x^{n-1}+\dots+x+1)\\ &=(x-1)(x^n+x^{n-1}+\dots+x+1).\end{align*}