8.1 Polynomial arithmetic and division

Why polynomials need their own arithmetic

Dividing $x^4-3x^3+2x^2+4x-1$ by $x^2-2x+3$ need not produce a polynomial quotient with no remainder. To imitate integer division, we need a condition that specifies when the remainder is small enough. For polynomials that condition concerns degree, not the value at a particular input: subtracting a suitable multiple of the divisor cancels the highest remaining power of $x$.

Throughout this chapter, the coefficients are real unless another field is explicitly named. The same definitions work over any field $F$, for example $Q$, $R$, or $C$.

Polynomials as finite formal sums

Definition

Polynomial over R

A polynomial with real coefficients is a formal sum

$$p(x)=\sum_{i=0}^{\infty} a_i x^i$$

where each $a_i\in R$ and all but finitely many coefficients are zero. The set of all such polynomials is denoted by $R[x]$.

The word "formal" matters. A polynomial is not merely a function value at one particular $x$; it is the whole list of coefficients, with only finitely many nonzero entries. Once the coefficients are fixed, the polynomial expression is fixed.

Concept lensStructural

Coefficients first, values after substitution

Two formal polynomials are equal precisely when every corresponding coefficient agrees, including omitted zero coefficients. Thus $1+x$ and $1+x+0x^2$ are equal. Evaluating at a real number $t$ instead produces the number $p(t)=\sum_i a_it^i$. The infinite summation notation involves no convergence question: only finitely many summands are nonzero.

An evaluated value contains less information than the coefficient list. For example, $x$ and $x^2$ both give $0$ at $t=0$ but have different coefficients. Over $\mathbb R$, agreement at every real input does imply polynomial equality; later in this note the root bound will justify that converse. Until then, coefficient equality is the definition, and evaluation is an operation applied to the already-defined object.

If at least one coefficient is nonzero, the degree of $p(x)$ is the largest index $i$ for which $a_i\ne0$. If every coefficient is zero, we call $p(x)$ the zero polynomial and set

$$\deg(0)=-\infty.$$

This convention lets degree formulas include the zero polynomial without many exceptional cases.

A nonzero constant has degree $0$, since its constant coefficient is its highest nonzero coefficient. The zero polynomial has no such coefficient and is not a degree-zero polynomial. In degree comparisons $-\infty$ is smaller than every nonnegative integer; use $\max(-\infty,m)=m$ and $(-\infty)+m=-\infty$, including $m=-\infty$. These are conventions for degree bookkeeping, not claims that a polynomial has a negative integer exponent. The zero polynomial has no leading coefficient and is not monic.

A polynomial is monic if its leading coefficient is $1$. For instance, $x^3-4x+7$ is monic, while $2x^3-4x+7$ is not.

Common mistake

Do not read the displayed last term as the degree

If a polynomial is written as

$$p(x)=a_0+a_1x+\cdots+a_nx^n,$$

then $\deg(p)=n$ only when $a_n\ne0$. The notation by itself does not guarantee that the displayed final coefficient is nonzero.

Addition, multiplication, and degree

Let

$$p(x)=\sum_{i=0}^{\infty}a_ix^i,\qquad q(x)=\sum_{i=0}^{\infty}b_ix^i.$$

Their sum is formed coefficient-by-coefficient:

$$p(x)+q(x)=\sum_{i=0}^{\infty}(a_i+b_i)x^i.$$

Their product is given by the convolution rule:

$$p(x)q(x)=\sum_{i=0}^{\infty}d_ix^i,\qquad d_i=\sum_{k=0}^{i}a_kb_{i-k}.$$

Only finitely many terms contribute, so the product is again a polynomial.

Theorem

Degree rules

For $p(x),q(x)\in R[x]$,

1. $\deg(p+q)\le \max\{\deg p,\deg q\}$;
2. $\deg(pq)=\deg p+\deg q$.

The second identity uses the fact that the coefficient field has no zero divisors.

The inequality in the first rule can be strict because leading terms may cancel. For example,

$$(x^2+1)+(-x^2+x)=x+1.$$

The product rule is stronger: if $p$ and $q$ are nonzero with leading coefficients $a_r$ and $b_s$, then the coefficient of $x^{r+s}$ in $pq$ is $a_rb_s$, which is nonzero.

To see why no higher product term survives, let $\deg p=r$ and $\deg q=s$. A contribution $a_kb_{i-k}$ at an index $i>r+s$ must have either $k>r$ or $i-k>s$, so one coefficient is zero. At index $r+s$, only the pair $k=r$, $i-k=s$ can contribute. Its product is nonzero because the coefficients belong to a field. Consequently a product of two nonzero polynomials cannot be the zero polynomial.

For addition, every coefficient above $\max(r,s)$ is zero. If the degrees are unequal, the higher leading term has no counterpart to cancel, so the bound is attained. With equal degrees, cancellation can continue through all coefficients: $p+(-p)=0$ has degree $-\infty$. If either factor in a product is zero, the product is zero and the stated conventions give the same rule.

Division by leading-term cancellation

Worked example

Divide a quartic by a quadratic

Find the quotient and remainder when

$$f(x)=x^4-3x^3+2x^2+4x-1$$

is divided by

$$g(x)=x^2-2x+3.$$

The first quotient term must cancel $x^4$. Comparing leading terms gives

$$\frac{x^4}{x^2}=x^2,$$

so subtract $x^2g$. Its leading term disappears, leaving a cubic. At each subsequent step, cancel the new leading term in the same way, retaining the entire remaining polynomial, including the constant term:

$$\begin{aligned} f-x^2g&=-x^3-x^2+4x-1,\\ (f-x^2g)-(-x)g&=-3x^2+7x-1,\\ (-3x^2+7x-1)-(-3)g&=x+8. \end{aligned}$$

After the first subtraction, the next leading-term comparison is

$$\frac{-x^3}{x^2}=-x,$$

and after the second it is

$$\frac{-3x^2}{x^2}=-3.$$

The remainder $x+8$ has degree $1$, below the divisor's degree $2$, so the process stops. The quotient accumulates the multipliers $x^2,-x,-3$.

In the second step, subtracting $(-x)g$ changes the signs of all its terms, not only the leading one. Keeping the unchanged constant $-1$ visible prevents a dropped term.

Long division gives

$$q(x)=x^2-x-3,\qquad r(x)=x+8.$$

Therefore

$$x^4-3x^3+2x^2+4x-1 =(x^2-2x+3)(x^2-x-3)+(x+8).$$

Follow the division steps: cancel each leading term to build the quotient, keeping $f=gq+r$ with $r$ as the current remainder.

1. Division identity

   For nonzero $g$, polynomial division writes $f(x)=g(x)q(x)+r(x)$ with $r=0$ or $\deg r < \deg g$.

2. Chapter example

   Divide $f=x^4-3x^3+2x^2+4x-1$ by $g=x^2-2x+3$, building $q$ and $r$ one leading term at a time.

3. Cancel $x^4$

   The leading ratio $x^4/x^2$ gives the first quotient term $x^2$; subtracting $x^2g$ leaves $-x^3-x^2+4x-1$.

4. Cancel $-x^3$

   Repeating the rule gives the second quotient term $-x$ and updates the current remainder to $-3x^2+7x-1$.

5. Stop by degree

   The last quotient term is $-3$; the remainder $x+8$ has degree $1$, which is smaller than $\deg(g)=2$.

6. Invariant

   The final identity is $f=(x^2-2x+3)(x^2-x-3)+(x+8)$.

Polynomial long division repeatedly cancels the current leading term while preserving $f=gq+r$, where $r$ is the current remainder. The process stops when the remainder is zero or its degree is smaller than the divisor degree.

Existence and uniqueness of quotient and remainder

The calculation works because each subtraction lowers the degree. The same cancellation proves that division with remainder is possible for every nonzero divisor; the degree bound also makes its result unique.

Theorem

Division algorithm for polynomials

Let $f(x),g(x)\in R[x]$ with $g(x)\ne0$. Then there exist unique polynomials $q(x),r(x)\in R[x]$ such that

$$f(x)=g(x)q(x)+r(x),\qquad \deg r\lt\deg g.$$

For existence, consider the nonempty set

$$S=\{f-gs:s\in\mathbb R[x]\}.$$

It contains $f$ by taking $s=0$. If $0\in S$, choose $q$ with $f-gq=0$ and take $r=0$; the remainder condition follows from $\deg0=-\infty$. Otherwise every element of $S$ has nonnegative integer degree, so well-ordering gives an element $r_0=f-gq_0$ of least degree.

Suppose $\deg r_0=k\ge j=\deg g$, with leading coefficients $c_k$ and $b_j$ respectively. Since $b_j\ne0$, the field permits the coefficient $c_k/b_j$. Form

$$r_1=r_0-g\left(\frac{c_k}{b_j}x^{k-j}\right) =f-g\left(q_0+\frac{c_k}{b_j}x^{k-j}\right).$$

The exponent $k-j$ is nonnegative, so the added expression is a polynomial and $r_1$ is still in $S$. The two leading terms $c_kx^k$ cancel and all remaining terms have degree below $k$. If $r_1=0$, it contradicts $0\notin S$; otherwise its smaller degree contradicts minimality. Therefore $\deg r_0\lt\deg g$, and $q_0,r_0$ supply the required pair.

Uniqueness is just as important as existence. If

$$f=gq_1+r_1=gq_2+r_2$$

with both remainders of degree less than $\deg g$, then

$$g(q_1-q_2)=r_2-r_1.$$

The left side is either zero or has degree at least $\deg g$; the right side has degree strictly less than $\deg g$. Thus both sides must be zero, so $q_1=q_2$ and $r_1=r_2$.

Proof X-Ray

Where the uniqueness contradiction occurs

Assume the quotients differ. Their difference is then a nonzero polynomial, whose degree is at least $0$. Multiplying by nonzero $g$ makes the left side have degree at least $\deg g$. On the right, subtraction may cancel terms, but cannot increase the degree beyond that of the larger remainder. Thus the two sides cannot be equal. The quotients must agree, and substituting back forces the remainders to agree too. This argument uses the degree bound and the nonzero-divisor hypothesis together.

What the degree condition permits

If $f=0$ and $g\ne0$, the unique pair is $q=r=0$. If $f\ne0$ but $\deg f\lt\deg g$, the unique pair is $q=0$, $r=f$; no cancellation step is required. If the divisor is a nonzero constant $c$, the remainder must have degree below $0$. Only the zero polynomial qualifies, so $q=f/c$ and $r=0$. Calling the zero polynomial degree $0$ would incorrectly exclude this valid remainder.

During long division, every unfinished nonzero remainder has a nonnegative integer degree. Cancelling its leading term strictly decreases that degree, so the process terminates. The final remainder need not be positive: unlike integer remainders, polynomials here have no sign restriction. A smaller degree, rather than a smaller value at a chosen input, is the stopping criterion.

Remainders from evaluation

When the divisor is linear, the division algorithm becomes especially powerful.

Theorem

Remainder theorem

Let $f(x)\in R[x]$ and $a\in R$. When $f(x)$ is divided by $x-a$, the remainder is $f(a)$.

Indeed, the remainder has degree less than $1$, so it is a constant $R$. Writing

$$f(x)=(x-a)q(x)+R$$

and substituting $x=a$ gives $f(a)=R$.

Here $R=f(a)$ is regarded as a constant polynomial in the division identity; it may be zero. We do not divide the identity by $x-a$ and then substitute $x=a$, since that would divide by zero. Evaluation is legitimate directly in the polynomial identity.

Polynomial divisibility means that $g\mid f$ if $f=gq$ for some polynomial $q$ in the same coefficient field. For nonzero $g$, this is equivalent to having zero remainder. In particular, every polynomial divides the zero polynomial, although division with remainder requires a nonzero divisor.

Theorem

Factor theorem

For $f(x)\in R[x]$ and $a\in R$,

$$(x-a)\mid f(x)\quad\Longleftrightarrow\quad f(a)=0.$$

The factor theorem translates between algebraic factorization and roots. It is one of the main reasons roots become useful: a root supplies a linear factor, and a linear factor supplies a root.

Worked example

Remainder modulo $x^2-1$

Suppose the remainders when $f(x)$ is divided by $x-1$ and $x+1$ are $5$ and $3$, respectively. Find the remainder when $f(x)$ is divided by $x^2-1$.

By the remainder theorem,

$$f(1)=5,\qquad f(-1)=3.$$

The remainder upon division by $x^2-1$ has degree less than $2$, so write it as $ax+b$. Then

$$a+b=5,\qquad -a+b=3.$$

Solving gives $a=1$ and $b=4$. The remainder is therefore

$$x+4.$$

Worked example

Choose a parameter with the factor theorem

Find $k$ so that

$$x-3$$

divides

$$f(x)=x^3+kx^2-4x+6.$$

By the factor theorem, $x-3$ divides $f(x)$ exactly when $f(3)=0$. Compute

$$f(3)=27+9k-12+6=21+9k.$$

Thus $21+9k=0$, so

$$k=-\frac{21}{9}=-\frac73.$$

The important point is that we did not divide the cubic by $x-3$. The factor theorem converts the factor condition into a single evaluation equation.

Checkpoint

What is the remainder when $f(x)=x^3+2x-5$ is divided by $x-2$?

Use the remainder theorem.

Solution · Answer

The remainder is $f(2)=8+4-5=7$.

How many roots can a nonzero polynomial have?

The factor theorem gives a degree bound on roots.

Theorem

Root bound

A nonzero polynomial of degree $n$ over $R$ or $C$ has at most $n$ distinct roots.

The base case is degree $0$: a nonzero constant never evaluates to zero. Assume the bound for degree $n$, and let $f$ have degree $n+1$. If it has no roots, the claim already holds. Otherwise choose a root $a$. The factor theorem gives $f(x)=(x-a)q(x)$, where $q$ is nonzero and has degree $n$ by the product rule.

For any distinct root $b\ne a$, evaluation gives $0=(b-a)q(b)$. The scalar $b-a$ is nonzero in the coefficient field, so $q(b)=0$. By induction there are at most $n$ such distinct roots, plus the one value $a$. This proves the bound for degree $n+1$. The argument does not require $q(a)\ne0$: even if $a$ is also a root of $q$, it is still only one value in the set of roots.

An immediate consequence is that a polynomial of degree at most $n$ with $n+1$ distinct roots must be the zero polynomial. This is a uniqueness theorem: too many zeros force every coefficient to vanish.

The root bound also needs its exact hypotheses. The zero polynomial vanishes at every input and is excluded. The word “distinct” counts values, rather than the number of times a factor occurs. Applying the bound to $p-q$ shows that polynomials of degree at most $n$ agreeing at $n+1$ distinct inputs are equal: otherwise their nonzero difference would have too many roots. Agreement at every real input is a special case, completing the connection between formal polynomials and their evaluated functions.

Common mistakes

Common mistake

Confusing equality of values with equality of polynomials

Checking that two polynomials agree at one value of $x$ does not prove they are the same polynomial. To prove equality of polynomials, either compare all coefficients or show that their difference has more roots than its degree allows.

Common mistake

Forgetting the degree condition on the remainder

An identity of the form $f=gq+r$ is not yet the division algorithm unless $\deg r\lt\deg g$. Without that condition, the quotient and remainder are not unique; one can move a multiple of $g$ back and forth between $q$ and $r$.

Summary

A polynomial is a finite-support formal sum. Degree measures the leading nonzero coefficient, and it controls addition, multiplication, and division. The division algorithm gives unique $q$ and $r$; the remainder theorem turns division by $x-a$ into evaluation at $a$; the factor theorem identifies roots with linear factors; and the root bound explains why too many roots force a polynomial to be zero.

Choosing an argument

For a degree calculation, check whether leading terms cancel before using the sum rule: two fourth-degree polynomials can sum to a quadratic, a linear polynomial, a constant, or zero. The product rule is exact for nonzero factors because their leading coefficients have nonzero product.

For the roots-of-unity exercise, use the factor theorem at $\omega$ and $\omega^2$, then use $\omega^3=1$. The divisibility of $F(x)+xG(x)$ becomes two linear equations in $f(1)$ and $g(1)$, without expanding a large polynomial.

Quick checks

Checkpoint

Why is the degree of the zero polynomial set to $-\infty$?

Think about formulas involving addition and multiplication.

Solution · Answer

The convention lets degree rules such as $\deg(0\cdot p)=\deg 0+\deg p$ remain formally consistent.

Checkpoint

If a nonzero polynomial has degree at most $4$, how many distinct roots can it have?

Use the root bound.

Solution · Answer

It can have at most $4$ distinct roots.

Exercises

1. Let $p(x)=3x^4-x^2+2$ and $q(x)=-3x^4+5x+1$. Find the upper bound for the degree of $p+q$ supplied by the degree rule, then compute its actual degree.
2. Divide $x^4-3x^3+2x^2+4x-1$ by $x^2-2x+3$.
3. Use the remainder theorem to find the remainder when $x^5-2x^2+7$ is divided by $x+1$.
4. Suppose $f(1)=5$ and $f(-1)=3$. Reconstruct the remainder of $f$ modulo $x^2-1$.
5. Prove that substitution preserves this divisibility relation: if $F(x)=f(x^3)$, $G(x)=g(x^3)$, and $F(x)+xG(x)$ is divisible by $x^2+x+1$, then both $f(x)$ and $g(x)$ are divisible by $x-1$.

Solution · Model solution 1

The degree rule gives the upper bound $4$, but the $3x^4$ and $-3x^4$ terms cancel. Thus $p+q=-x^2+5x+3$, whose degree is $2$.

Solution · Model solution 2

The quotient is $x^2-x-3$ and the remainder is $x+8$.

Solution · Model solution 3

Since the divisor is $x-(-1)$, the remainder is $(-1)^5-2(-1)^2+7=-1-2+7=4$.

Solution · Model solution 4

Write the remainder as $ax+b$. Then $a+b=5$ and $-a+b=3$, so $a=1$, $b=4$, and the remainder is $x+4$.

Solution · Model solution 5

Let $\omega=e^{2\pi i/3}$. Since $x^2+x+1=(x-\omega)(x-\omega^2)$, the hypothesis gives $f(1)+\omega g(1)=0$ and $f(1)+\omega^2g(1)=0$, because $\omega^3=(\omega^2)^3=1$. Subtracting gives $(\omega-\omega^2)g(1)=0$, and $\omega\ne\omega^2$, hence $g(1)=0$, and then $f(1)=0$. By the factor theorem, $x-1$ divides both $f(x)$ and $g(x)$.