Polynomial Factor Theorem

Theorem
Let $P \left({x}\right)$ be a polynomial in $x$ over a field $K$ of degree $n$.

Then $\xi \in K: P \left({\xi}\right) = 0$ if and only if $P \left({x}\right) = \left({x - \xi}\right) Q \left({x}\right)$, 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 $P \left({\xi_1}\right) = P \left({\xi_2}\right) = \ldots = P \left({\xi_n}\right) = 0$, then:
 * $\displaystyle P \left({x}\right) = k \prod_{j=1}^n \left({x - \xi_j}\right)$

where $k \in K$.

Corollary
Let $P \left({x}\right)$ be a polynomial in $x$ over the real numbers $\R$ of degree $n$.

Suppose there exists $\xi \in \R: P \left({\xi}\right) = 0$.

Then $P \left({x}\right) = \left({x - \xi}\right) Q \left({x}\right)$, where $Q \left({x}\right)$ is a polynomial of degree $n - 1$.

Hence, if $\xi_1, \xi_2, \ldots, \xi_n \in \R$ such that all are different, and $P \left({\xi_1}\right) = P \left({\xi_2}\right) = \ldots = P \left({\xi_n}\right)$, then:
 * $\displaystyle P \left({x}\right) = k \prod_{j=1}^n \left({x - \xi_j}\right)$

where $k \in \R$.

Proof
Clearly if $P = (x-\xi)Q$ then $P(\xi) = Q(\xi) \cdot 0 = 0$.

Conversely suppose that $P(\xi) = 0$.

By the division theorem for polynomials there exist polynomials $Q$ and $R$ such that $P = Q(X-\xi) + R$ and $\operatorname{deg}R < \operatorname{deg}(X-\xi) = 1$.

Evaluating at $\xi$ we have $0 = P(\xi) = R(\xi)$.

But $\operatorname{deg}R = 0$, so $R \in K$, in particular $R = 0$.

Thus $P = Q(X-\xi)$ as required.

The fact that $\operatorname{deg}Q = n-1$ follows from the fact that the ring of polynomials is an integral domain and the properties of degree of product of polynomials.

We can then apply this result to the situation where $P \left({\xi_1}\right) = P \left({\xi_2}\right) = \ldots = P \left({\xi_n}\right)$.

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 $x$, we get that $0 = P \left({\xi_2}\right) = \left({\xi_2 - \xi_1}\right) Q_{n-1} \left({x}\right)$.

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.

The result follows.

Proof of Corollary
Follows directly from the fact that the real numbers $\R$ form a field.