Polynomial Factor Theorem

Theorem
Let $P \paren x$ be a polynomial in $x$ over a field $K$ of degree $n$.

Then:
 * $\xi \in K: P \paren \xi = 0 \iff P \paren x = \paren {x - \xi} Q \paren x$

where $Q$ is a polynomial of degree $n - 1$.

Hence, if $\xi_1, \xi_2, \ldots, \xi_n \in K$ such that all are different, and $P \paren {\xi_1} = P \paren {\xi_2} = \ldots = P \paren {\xi_n} = 0$, then:
 * $\displaystyle P \paren x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$

where $k \in K$.

Proof
Let $P = \left({x - \xi}\right) Q$.

Then:
 * $P \left({\xi}\right) = Q \left({\xi}\right) \cdot 0 = 0$

Conversely, let $P \left({\xi}\right) = 0$.

By the Division Theorem for Polynomial Forms over Field, there exist polynomials $Q$ and $R$ such that:
 * $P = Q \left({x - \xi}\right) + R$

and:
 * $\deg R < \deg \left({x - \xi}\right) = 1$

Evaluating at $\xi$ we have:
 * $0 = P \left({\xi}\right) = R \left({\xi}\right)$

But:
 * $\deg R = 0$

so:
 * $R \in K$

In particular:
 * $R = 0$

Thus:
 * $P = Q \left({x - \xi}\right)$

as required.

The fact that $\deg Q = n - 1$ follows from:
 * Ring of Polynomial Forms is Integral Domain

and:
 * Degree of Product of Polynomials over Integral Domain.

We can then apply this result to:
 * $P \left({\xi_1}\right) = P \left({\xi_2}\right) = \ldots = P \left({\xi_n}\right) = 0$

We can progressively work through:
 * $P \left({x}\right) = \left({x - \xi_1}\right) Q_{n-1} \left({x}\right)$

where $Q_{n-1} \left({x}\right)$ is a polynomial of order $n - 1$.

Then, substituting $\xi_2$ for $x$:
 * $0 = P \left({\xi_2}\right) = \left({\xi_2 - \xi_1}\right) Q_{n-1} \left({x}\right)$

Since $\xi_2 \ne \xi_1$:
 * $Q_{n-1} \left({\xi_2}\right) = 0$

and we can apply the above result again:


 * $Q_{n-1} \left({x}\right) = \left({x - \xi_2}\right) Q_{n-2} \left({x}\right)$

Thus
 * $P \left({x}\right) = \left({x - \xi_1}\right) \left({x - \xi_2}\right) Q_{n-2} \left({x}\right)$

and we then move on to consider $\xi_3$.

Eventually we reach:
 * $P \left({x}\right) = \left({x - \xi_1}\right) \left({x - \xi_2}\right) \ldots \left({x - \xi_n}\right) Q_0 \left({x}\right)$

$Q_0 \left({x}\right)$ is a polynomial of zero degree, that is a constant polynomial.

The result follows.