Polynomial Factor Theorem
Theorem
Let $\map P x$ be a polynomial in $x$ over a field $K$ of degree $n$.
Then:
- $\xi \in K: \map P \xi = 0 \iff \map P x = \paren {x - \xi} \map Q x$
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:
- $\ds \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
where $k \in K$.
Corollary
Let $\map P x$ be a polynomial in $x$ over the real numbers $\R$ of degree $n$.
Suppose there exists $\xi \in \R: \map P \xi = 0$.
Then $\map P x = \paren {x - \xi} \map Q x$, where $\map Q x$ is a polynomial of degree $n - 1$.
Hence, if $\xi_1, \xi_2, \dotsc, \xi_n \in \R$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
- $\ds \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
where $k \in \R$.
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:
and:
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.
$\blacksquare$
Also known as
Some sources refer to the Polynomial Factor Theorem (and its variants) as the Factor Theorem, but there are multiple theorems with this name.
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $24$