[proofplan] We prove the equivalence by applying the division algorithm in $k[x]$ to divide $f$ by the monic linear polynomial $(x - a)$, producing a quotient and a constant remainder. Evaluating at $a$ shows the remainder equals $f(a)$, which establishes both directions of the if-and-only-if. The bound on roots then follows by induction on the degree, using the fact that each root produces a linear factor that strictly reduces the degree. [/proofplan] [step:Divide $f$ by $(x - a)$ using the division algorithm in $k[x]$] Since $k$ is a field, $k[x]$ is a Euclidean domain with the degree function as the Euclidean norm. Apply the [Division Algorithm for Polynomials](/theorems/???): since $(x - a)$ is a nonzero polynomial of degree $1$, there exist unique $q \in k[x]$ and $r \in k[x]$ with \begin{align*} f = (x - a)\, q + r, \qquad \deg(r) < \deg(x - a) = 1. \end{align*} The condition $\deg(r) < 1$ forces $\deg(r) \leq 0$, so $r \in k$ is a constant polynomial. [guided] Our goal is to relate divisibility of $f$ by $(x - a)$ to the value $f(a)$. The natural tool is polynomial long division. Since $k$ is a field, $k[x]$ is a Euclidean domain, and the division algorithm guarantees that for any nonzero divisor we can write the dividend as divisor times quotient plus remainder, with the remainder having strictly smaller degree. Apply the [Division Algorithm for Polynomials](/theorems/???): the divisor $(x - a)$ has degree $1$ and is nonzero, so there exist unique $q \in k[x]$ and $r \in k[x]$ satisfying \begin{align*} f = (x - a)\, q + r, \qquad \deg(r) < \deg(x - a) = 1. \end{align*} The constraint $\deg(r) < 1$ means $r$ has degree $0$ or $r = 0$, i.e., $r \in k$ is a constant. This constant remainder is the key object: we will identify it with $f(a)$. [/guided] [/step] [step:Evaluate at $x = a$ to identify the remainder as $f(a)$] Substituting $x = a$ into the identity $f = (x - a)\, q + r$ gives \begin{align*} f(a) = (a - a)\, q(a) + r = 0 \cdot q(a) + r = r. \end{align*} Therefore $r = f(a)$, and the division identity becomes \begin{align*} f = (x - a)\, q + f(a). \end{align*} [guided] The identity $f = (x - a)\, q + r$ holds as an equality in $k[x]$, and therefore as an equality of polynomial functions on $k$. Evaluating both sides at $x = a$: \begin{align*} f(a) = (a - a)\, q(a) + r = 0 \cdot q(a) + r = r. \end{align*} So the constant remainder $r$ is nothing other than $f(a)$. This is the heart of the argument: the value of $f$ at $a$ is the remainder upon division by $(x - a)$. Rewriting: \begin{align*} f = (x - a)\, q + f(a). \end{align*} This single identity immediately gives both directions of the equivalence. [/guided] [/step] [step:Conclude the equivalence: $(x - a) \mid f$ if and only if $f(a) = 0$] From $f = (x - a)\, q + f(a)$: **Forward direction.** Suppose $(x - a) \mid f$. Then $f = (x - a)\, h$ for some $h \in k[x]$. By uniqueness of the quotient and remainder in the division algorithm, $r = 0$, so $f(a) = r = 0$. **Reverse direction.** Suppose $f(a) = 0$. Then $f = (x - a)\, q + 0 = (x - a)\, q$, so $(x - a) \mid f$ in $k[x]$. [guided] The identity $f = (x - a)\, q + f(a)$ encodes both directions of the equivalence. We read them off one at a time. **Forward direction.** Assume $(x - a) \mid f$, i.e., there exists $h \in k[x]$ with $f = (x - a)\, h$. This expression is also a valid division-algorithm representation: $f = (x - a)\, h + 0$, with quotient $h$ and remainder $0$. By the uniqueness clause of the [Division Algorithm for Polynomials](/theorems/???) — which guarantees that the quotient and remainder are uniquely determined — the remainder must equal the one we already found, namely $f(a)$. Therefore $f(a) = 0$. **Reverse direction.** Assume $f(a) = 0$. Substituting into the division identity gives $f = (x - a)\, q + 0 = (x - a)\, q$, which is precisely the statement that $(x - a)$ divides $f$ in the polynomial ring $k[x]$, with quotient $q$. Notice that both directions ultimately rest on the same mechanism: the division algorithm produces a unique remainder, and that remainder is $f(a)$. The forward direction uses uniqueness to force the remainder to be zero; the reverse direction uses the identity directly. Why does the proof require $k$ to be a field? The division algorithm in $k[x]$ requires that the leading coefficient of the divisor be a unit — which is automatic when $k$ is a field (every nonzero element is a unit). Over a general commutative ring, division with remainder may not exist, and the Factor Theorem can fail. [/guided] [/step] [step:Bound the number of roots by induction on $\deg(f)$] We prove by induction on $n = \deg(f)$ that a nonzero polynomial $f \in k[x]$ of degree $n$ has at most $n$ roots in $k$, counted with multiplicity. **Base case ($n = 0$).** A nonzero constant polynomial $f \in k \setminus \{0\}$ satisfies $f(a) \neq 0$ for all $a \in k$, so $f$ has $0$ roots, and $0 \leq 0$. **Inductive step.** Let $n \geq 1$ and assume every nonzero polynomial in $k[x]$ of degree less than $n$ has at most as many roots as its degree. Suppose $f \in k[x]$ has degree $n$. If $f$ has no roots in $k$, the bound $0 \leq n$ holds. Otherwise, let $a \in k$ be a root of $f$, so $f(a) = 0$. By the equivalence proved above, $(x - a) \mid f$, so $f = (x - a)\, q$ for some $q \in k[x]$. Comparing degrees: $\deg(f) = \deg(x - a) + \deg(q) = 1 + \deg(q)$, so $\deg(q) = n - 1$. Since $k$ is a field, $k[x]$ is an [integral domain](/theorems/???): it has no zero divisors. Therefore, for any $b \in k$, \begin{align*} f(b) = (b - a)\, q(b) = 0 \quad \implies \quad b = a \;\text{ or }\; q(b) = 0. \end{align*} Every root of $f$ distinct from $a$ is a root of $q$. By the inductive hypothesis, $q$ has at most $n - 1$ roots in $k$ (counted with multiplicity). Including $a$, the polynomial $f$ has at most $1 + (n - 1) = n$ roots in $k$, counted with multiplicity. [guided] The root-count bound is a natural corollary of the Factor Theorem. We prove it by strong induction on the degree. **Base case ($n = 0$).** If $\deg(f) = 0$, then $f$ is a nonzero element of $k$, and evaluating $f$ at any $a \in k$ gives that same nonzero constant. So $f$ has zero roots, satisfying $0 \leq 0$. **Inductive step.** Let $n \geq 1$ and assume the statement holds for all nonzero polynomials of degree strictly less than $n$. Let $f \in k[x]$ have degree $n$. If $f$ has no roots in $k$, the bound is vacuously satisfied. So suppose $a \in k$ is a root: $f(a) = 0$. By the Factor Theorem (the equivalence just established), $(x - a) \mid f$, and we write $f = (x - a)\, q$ for some $q \in k[x]$. Since $\deg(f) = \deg(x - a) + \deg(q)$ (which holds because $k[x]$ is an integral domain — leading coefficients multiply without cancellation), we get $\deg(q) = n - 1$. Note $q$ is nonzero because $f$ is nonzero. Now consider any root $b \in k$ of $f$: \begin{align*} f(b) = (b - a)\, q(b) = 0. \end{align*} Since $k$ is a field, $k[x]$ is an [integral domain](/theorems/???), so the product $(b - a)\, q(b)$ vanishes only if $b - a = 0$ (i.e., $b = a$) or $q(b) = 0$. This is where the integral domain property is essential: over a ring with zero divisors (such as $\mathbb{Z}/6\mathbb{Z}$), a product can vanish without either factor vanishing, and the root-count bound fails. Every root of $f$ other than $a$ must be a root of $q$. By the inductive hypothesis applied to $q$ (which has degree $n - 1 < n$), the polynomial $q$ has at most $n - 1$ roots in $k$. Adding the root $a$, we find $f$ has at most $n$ roots in $k$, counted with multiplicity. This completes the induction. [/guided] [/step]