The Remainder Theorem for Polynomials over a Field
 Table of Contents

# The Remainder Theorem for Polynomials over a Field

 Definition: Let $(F, +, \cdot)$ be a field and let $f, g \in F[x]$. The polynomial $g$ is said to be a Divisor or Factor of $f$ is there exists a polynomial $q \in F[x]$ such that $f(x) = g(x)q(x)$ which we denote by $g(x) | f(x)$ or $g | f$.

The easiest examples come from the field of real numbers. Consider the polynomial:

(1)
\begin{align} \quad f(x) = x^2 - 9 \end{align}

Then $f(x)$ can factored as:

(2)
\begin{align} \quad f(x) = (x - 3)(x + 3) \end{align}

So $(x - 3)$ is a factor of $x^2 - 9$. Similarly, $(x + 3)$ is a factor of $x^2 - 9$.

 Lemma 1: Let $(F, +, \cdot)$ be a field. Then for all $c \in F$ and for all $n \in \mathbb{N}$ we have that $(x - c) | (x^n - c^n)$.
• Proof:
(3)
\begin{align} \quad x^n - c^n = (x - c)\underbrace{(x^{n-1} + cx^{n-2} + ... + c^{n-2}x + c^{n-1})}_{q(x)} \end{align}
• So $(x - c) | (x^n - c^n)$. $\blacksquare$

We use Lemma 1 above to prove the important Remainder theorem which tells us that given any field $F$, any nonzero polynomial $f$ over $F$, and any element $c$ in $F$ there exists a unique polynomial $q$ over $F$ such that $f$ is equal to the product of the linear term $(x - c)$ with $q$ plus $f$ evaluated at $c$.

 Theorem 1 (The Remainder Theorem): Let $(F, +, \cdot)$ be a field and let $f \in F[x]$ with $f(x) \neq 0$ and let $c \in F$. Then there exists a unique polynomial $q \in F[x]$ such that $f(x) = (x - c)q(x) + f(c)$.
• Proof: Let $f \in F[x]$ with $f(x) \neq 0$. Then $f(x)$ has the following form (where $a_n \neq 0$):
(4)
\begin{align} \quad f(x) = a_0 + a_1x + ... + a_nx^n \end{align}
• For $c \in F$ consider the difference $f(x) - f(c)$:
(5)
\begin{align} \quad f(x) - f(c) &= [a_0 + a_1x + ... + a_nx^n] - [a_0 + a_1c + ... + a_nc^n] \\ &= a_1(x - c) + ... + a_n(x^n - c^n) \end{align}
• From the previous lemma we see that $(x - c) | a_1(x - c) + ... + a_n(x^n - c^n)$ so there exists a $q \in F[x]$ such that:
(6)
\begin{align} \quad f(x) - f(c) &= (x - c)q(x) \\ \quad f(x) &= (x - c)q(x) + f(c) \end{align}
• We now show that $q \in F[x]$ is unique. Suppose there exists a $q' \in F[x]$ such that $f(x) = (x - c)q'(x) + f(c)$. Then:
(7)
\begin{align} \quad (x - c)q(x) &= (x - c)q'(x) \\ \quad (x - c)[q(x) - q'(x)] &= 0 \end{align}
• Since $(x - c) \neq 0$ the equality above implies that $q(x) - q'(x) = 0$ so $q(x) = q'(x)$. Thus $q \in F[x]$ is unique. $\blacksquare$
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