Polynomial Factor Theorem

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

Let $$\exist \xi \in \mathbb{R}: P \left({\xi}\right) = 0$$.

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

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

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

Then:

$$ $$ $$

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

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.