Polynomial Factor Theorem

Theorem
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
By the definition of a polynomial, let $\displaystyle P \left({x}\right) = \sum_{j=0}^n a_j x^j$.

Suppose $P \left({\xi}\right) = 0$.

Then:

But from Difference of Two Powers‎, for each of the above $x^j - \xi^j$ we have:
 * $\displaystyle x^j - \xi^j = \left({x - \xi}\right) \sum_{i=0}^{j-1} x^{j-i-1} \xi^i$

Thus $x^j - \xi^j = \left({x - \xi}\right) F_{j-1} \left({x}\right)$, where $F \left({x}\right)$ is a polynomial of degree $j - 1$.

Thus:
 * $P \left({x}\right) = \left({x - \xi}\right) \left({a_n F_{n-1} \left({x}\right) + a_{n-1} F_{n-2} \left({x}\right) + \ldots + a_1}\right)$

By Polynomials Closed under Addition, it follows that:
 * $a_n F_{n-1} \left({x}\right) + a_{n-1} F_{n-2} \left({x}\right) + \ldots + a_1 = Q \left({x}\right)$

is itself a polynomial of order $n-1$.

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

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


 * 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.