Factoring a polynomial over $\mathbb{C}$

Question

Every polynomial with complex coefficients can be written as the product of linear factors. What are the linear factors of $P(z)=1+z+⋯+z^7+z^8$?

Answer 1

Notice that $P(1)\neq 0$. Now let's multiply by $(z-1)$ to obtain $(z-1)P(z)=z^9-1$. Since $1$ is not a root of $P(z)$, the linear factors of this new polynomial are going to be exactly $(z-1)$ and then the linear factors of $P(z)$. Clearly the roots are the ninth roots of unity, so the linear factors of $P(z)$ are of the form $(z-\zeta)$ where $\zeta\neq 1$ is a ninth root of unity.

Answer 2

Do you know the geometric series formula? You can write

$$\frac{X^9 - 1}{X-1} = 1 + X + \cdots + X^8 = P(X)$$

Now the roots of $X^9 - 1$ are $e^{\frac{2 \pi i k}{9}}$ for $0 \leq k \leq 8$. Removing the root of $X - 1$ leaves you with $1 \leq k \leq 8$.