Division and Roots in F[x]

Long division of polynomials works over any field, because each step divides by a leading coefficient. That single fact gives the remainder theorem, the factor theorem and a bound on the number of roots.

Por qué funciona

Division algorithm. Let $F$ be a field and $g \ne 0$ in $F[x]$. For every $f$ there are unique $q, r$ with $$f = qg + r, \qquad r = 0 \text{ or } \deg r \lt \deg g.$$ Existence: while $\deg f \ge \deg g$, subtract $\frac{a}{b}x^{m - n}g$, where $ax^{m}$ and $bx^{n}$ are the leading terms; this kills the leading term of $f$ and lowers its degree. (Here $\frac{a}{b}$ means $a$ times the inverse of $b$, which exists in a field.) Uniqueness: if $qg + r = q'g + r'$, then $(q - q')g = r' - r$; the right side has degree less than $\deg g$, so both sides are $0$. $\square$

Remainder and factor theorems. Dividing by $x - c$ leaves the constant $f(c)$ (substitute $x = c$ into $f = q(x)(x - c) + r$). So $c$ is a root exactly when $x - c$ divides $f$, as in algebra 2. Root bound: a nonzero polynomial of degree $n$ over a field (or integral domain) has at most $n$ roots, because each root splits off a factor $x - c$ and degrees add.

Ejemplo resuelto: Dividing in $\mathbb{Z}_{3}[x]$

Divide $f = x^{3} + 2x + 1$ by $g = 2x + 1$ in $\mathbb{Z}_{3}[x]$.

The leading coefficient of $g$ is $2$, and $2^{-1} = 2$ in $\mathbb{Z}_{3}$. First quotient term: $x^{3} \cdot 2 / x = 2x^{2}$. Subtract $2x^{2}(2x + 1) = 4x^{3} + 2x^{2} = x^{3} + 2x^{2}$: left $-2x^{2} + 2x + 1 = x^{2} + 2x + 1$.
Next term: $x^{2} \cdot 2/x = 2x$. Subtract $2x(2x + 1) = x^{2} + 2x$: left $1$.
So $q = 2x^{2} + 2x$ and $r = 1$. Check: $(2x^{2} + 2x)(2x + 1) + 1 = 4x^{3} + 6x^{2} + 2x + 1 = x^{3} + 2x + 1$.

Respuesta: $q = 2x^{2} + 2x$, $r = 1$.

Ejemplo resuelto: Remainder without dividing

Find the remainder of $x^{4} + 2x^{2} + 5$ divided by $x - 3$ in $\mathbb{Z}_{7}[x]$.

Remainder $= f(3) = 81 + 18 + 5 = 104$.
$104 = 14 \cdot 7 + 6$.

Respuesta: $6$.

Ejemplo resuelto: Over $\mathbb{Q}$

Divide $2x^{3} - 3x^{2} + x + 5$ by $x - 2$ in $\mathbb{Q}[x]$.

$2x^{3} \div x = 2x^{2}$; subtract $2x^{3} - 4x^{2}$: left $x^{2} + x + 5$.
$x^{2} \div x = x$; subtract $x^{2} - 2x$: left $3x + 5$.
$3x \div x = 3$; subtract $3x - 6$: left $11$. Check: $f(2) = 16 - 12 + 2 + 5 = 11$.

Respuesta: $q = 2x^{2} + x + 3$, $r = 11$.

Error común

Dividing by a leading coefficient as if you were in $\mathbb{Q}$: in $\mathbb{Z}_{3}$, "divide by 2" means multiply by $2^{-1} = 2$, never write $\frac{1}{2}$. And the root bound needs a field (or integral domain): in $\mathbb{Z}_{8}$, $x^{2} - 1$ has the four roots $1, 3, 5, 7$, because $(x - 3)(x - 5) = x^{2} - 8x + 15 = x^{2} - 1$ there too.

Division finds factors. A zero remainder hands you the factorization. Over $\mathbb{Z}_{5}$, dividing $x^{3} + 1$ by $x + 1$ leaves $(-1)^{3} + 1 = 0$, and the quotient gives $x^{3} + 1 = (x + 1)(x^{2} + 4x + 1)$.

Idea clave

Over a field, $f = qg + r$ with $\deg r \lt \deg g$, uniquely. Dividing by $x - c$ leaves $f(c)$, so $c$ is a root iff $(x - c) \mid f$, and a degree-$n$ polynomial has at most $n$ roots.

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