Polynomial Factor Theorem

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

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

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 $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
 * $\displaystyle \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$

where $k \in K$.

Proof
Let $P = \paren {x - \xi} Q$.

Then:
 * $\map P \xi = \map Q \xi \cdot 0 = 0$

Conversely, let $\map P \xi = 0$.

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

and:
 * $\map \deg R < \map \deg {x - \xi} = 1$

Evaluating at $\xi$ we have:
 * $0 = \map P \xi = \map R \xi$

But:
 * $\deg R = 0$

so:
 * $R \in K$

In particular:
 * $R = 0$

Thus:
 * $P = \map Q {x - \xi}$

as required.

The fact that $\map \deg Q = n - 1$ 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:
 * $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$

We can progressively work through:
 * $\map P x = \paren {x - \xi_1} \map {Q_{n - 1} } x$

where $\map {Q_{n - 1} } x$ is a polynomial of order $n - 1$.

Then, substituting $\xi_2$ for $x$:
 * $0 = \map P {\xi_2} = \paren {\xi_2 - \xi_1} \map {Q_{n - 1} } x$

Since $\xi_2 \ne \xi_1$:
 * $\map {Q_{n - 1} } {\xi_2} = 0$

and we can apply the above result again:


 * $\map {Q_{n - 1} } x = \paren {x - \xi_2} \map {Q_{n - 2} } x$

Thus:
 * $\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \map {Q_{n - 2} } x$

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

Eventually we reach:
 * $\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \dotsm \paren {x - \xi_n} \map {Q_0} x$

$\map {Q_0} x$ is a polynomial of zero degree, that is a constant polynomial.

The result follows.