# Roots of Polynomials over a Field

Recall from The Remainder Theorem for Polynomials over a Field page that if $(F, +, \cdot)$ is a field and $f, g \in F[x]$ then we say that $g$ is a divisor or factor of $f$ is there exists a polynomial $q \in F[x]$ such that:

(1)We denote that $g(x) | f(x)$ or $g | f$.

We also proved a very nice theorem called the Remainder theorem which states that if $f \in F[x]$, $f(x) \neq 0$, and $c \in F$ then there exists a unique polynomial $q \in F[x]$ such that:

(2)We now move on to another important concept of polynomials - roots.

Definition: Let $(F, +, \cdot)$ be a field and let $f \in F[x]$. An element $c \in F$ is called a Root or Solution to $f$ is $f(c) = 0$. |

For example, consider the field $(\mathbb{Z}_3, +, \cdot)$ of integers modulo $3$ and the following polynomial:

(3)Then $2 \in \mathbb{Z}_3$ is a root of $f$ since:

(4)We now prove an important result connecting divisibility and roots which states that an element $c \in F$ is a root of $f$ if and only if the term $x - c$ is a divisor of $f(x)$.

Theorem 1: Let $(F, +, \cdot)$ be a field and let $f \in F[x]$ with $f(x) \neq 0$ and let $c \in F$. Then $c$ is a root of $f$ if and only if $(x - c) | f(x)$. |

**Proof:**$\Rightarrow$ Suppose that $c$ is a root of $f$. Then $f(c) = 0$. By the Remainder theorem we have that there exists a unique $q \in F[x]$ such that:

- So $(x - c) | f(x)$.

- $\Leftarrow$ Suppose that $(x - c) | f(x)$. Then there exists a $q \in F[x]$ such that $f(x) = (x - c)q(x)$. So $f(c) = (c - c)q(x) = 0q(x) = 0$. Hence $c$ is a root of $f$. $\blacksquare$

The next theorem tells us that if $f \in F[x]$ is a polynomial of degree $n$ then $f$ has at most $n$ distinct roots.

Theorem 2: Let $(F, +, \cdot)$ be a field and let $f \in F[x]$ with $f(x) \neq 0$. If $\deg (f) = n$ then $f$ has at most $n$ distinct roots. |

**Proof:**We carry this proof out by mathematical induction.

- If $f(x) = a_0$ with $a_0 \neq 0$ then $\deg (f) = 0$ and furthermore $f$ has no roots.

- If $f(x) = a_0 + a_1x$ with $a_1 \neq 0$ then $\deg(f) = 1$ and furthermore $f$ has exactly one root, namely $\displaystyle{x = \frac{-a_0}{a_1}}$.

- Assume that the theorem is true for all polynomials in $F[x]$ of degree $n - 1$ and let $f(x) = a_0 + a_1x + ... + a_nx^n$ with $a_n \neq 0$. Then $\deg (f) = n$ If $f$ has no roots then $f$ has at most $n$ roots. If $f$ has a root $c \in F$ then by the previous theorem there exists a $q \in F[x]$ such that:

- Here $\deg (q) = n - 1$. By the induction hypothesis $q$ has at most $n - 1$ roots. So $f$ has at most $n$ roots. $\blacksquare$