Show that no linear polynomial divides $x^k + x^{k-1} + \cdots + 1$ with $k\ge 2$ even

Question

Let $f(x) = x^k + x^{k-1} + \cdots+ 1 \in \mathbb{Q}[x]$, $k\ge 2$ and even. Show there's no linear polynomial which divides $f(x)$.

A start:
Lets assume by contradiction there's a linear polynomial $ax+b$ such which divides $f(x)$.
Therefore, for some $q(x)$: $f(x) = q(x)(ax+b)$. Hence, $-\frac{b}{a}$ is a root of $f(x)$.

$$f\left(\frac{-b}{a}\right) = \left(\frac{-b}{a}\right)^k + \left(\frac{-b}{a}\right)^{k-1} + \cdots+1 = 0$$

What should I do next?

Answer 1

Use the rational root theorem. It says that if $\frac pq$ is a root (in the lowest terms), then $p$ divides $a_0$ and $q$ divides $a_n$, so the roots can be only $\pm1$.

Answer 2

You are right in that $f$ must have a root. Call it $u$. We know that $u\neq 1$ because $f(1)=k+1\neq 0$.

Then, $$u^k+u^{k-1}+\cdots+1=\sum_{j=0}^k u^k=0$$

But for $u\neq 1$ this is the same as $$\frac{u^{k+1}-1}{u-1}=0$$

But this is not possible for $u\neq 1$.