8.1 Polynomial arithmetic and division

Why polynomials need their own arithmetic

A polynomial is familiar as an expression such as $x^4-3x^3+2x^2+4x-1$, but chapter 8 treats polynomials more carefully than ordinary algebraic shorthand. The point is not only to manipulate expressions. The point is to build an arithmetic system that behaves enough like the integers to support division with remainder, greatest common divisors, factorization, and later partial fraction decompositions.

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

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.

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

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.

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.

The division algorithm

The central structural result is the polynomial version of integer division. It says that when we divide by a nonzero polynomial, there is a unique quotient and a unique remainder whose degree is smaller than the divisor.

The proof follows a minimal-degree idea. Consider all possible expressions $f(x)-g(x)s(x)$ as s(x) ranges over R[x]. If zero appears, the remainder is zero. Otherwise choose an element of least degree. If its degree were still at least $\deg g$, subtract a suitable multiple of g(x) that cancels its leading term. That creates an even smaller-degree element, a contradiction.

Uniqueness is just as important as existence. If

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

with both remainders smaller than 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 $r_1=r_2$.

Remainders from evaluation

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

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 $f(a)=R$.

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.

How many roots can a nonzero polynomial have?

The factor theorem gives a degree bound on roots.

The proof is by induction on degree. If f has a root a, then $f(x)=(x-a)q(x)$ and $\deg q=\deg f-1$. Every root of f other than a is a root of q. The induction hypothesis then gives the bound.

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.

Proof sketch or proof idea

The division algorithm is often introduced as a procedure, but in this course it is also a theorem with existence and uniqueness content. The existence part does not depend on the particular layout of long division. It depends on a simple descent principle for degrees. Among all expressions of the form $f-gs$, choose one with minimal degree. If its degree is still at least the degree of g, then its leading term can be cancelled by subtracting a suitable multiple of g. That cancellation produces another element of the same set with smaller degree, contradicting the choice of a minimal-degree remainder.

The uniqueness part is equally structural. If two quotient-remainder pairs worked, subtracting the two identities would say that a multiple of g equals the difference of two small remainders. A nonzero multiple of g cannot have degree smaller than g; the difference of two allowed remainders must have degree smaller than g. The only way out is that both sides are zero. This is why the phrase "the quotient and the remainder" is justified.

This proof pattern is worth remembering because it reappears in the next note. The Euclidean algorithm for polynomial gcds works only because every division step produces a strictly smaller-degree remainder. Degree plays the role that absolute value played for positive integers: it is the quantity that must descend and therefore cannot descend forever.

How to read a polynomial long division

A long division table should not be read as a mysterious arrangement of symbols. It is a repeated leading-term cancellation algorithm.

Start with the current dividend. Compare its leading term with the leading term of the divisor. In Example 8.0, the first comparison is

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

That quotient term is chosen for one reason only: multiplying the divisor by $x^2$ creates a leading term $x^4$, which cancels the current leading term. After subtraction, the current remainder becomes $-x^3-x^2+4x-1$. The same logic gives the next term

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

and then the final quotient term

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

The algorithm stops not because the expression looks simpler, but because the current remainder has degree 1, which is smaller than the divisor degree 2. This stopping condition is exactly the condition in the theorem. If a student keeps dividing after the remainder has smaller degree, they are no longer following the division algorithm.

Common mistakes

Summary

This note sets up the algebraic foundation for the rest of chapter 8. 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.

Study guide for the exercises

When solving the exercises, separate three tasks that are easy to mix together. First, simplify the expression only by operations that preserve the polynomial identity being studied. Second, keep track of the degree condition whenever a remainder appears. Third, decide whether the problem is asking for a calculation or for a proof about all polynomials of a given form.

For degree questions, always inspect possible cancellation before announcing the degree of a sum. The degree rule for sums gives only an upper bound. A sum of two fourth-degree polynomials may become quadratic, linear, constant, or even zero if leading terms cancel. For products, the leading coefficient argument is stronger, so the degree is exactly additive as long as both factors are nonzero.

For remainder and factor theorem questions, resist doing unnecessary long division. If the divisor is $x-a$, evaluation at a is the direct route. If the divisor is a product such as $(x-1)(x+1)$, the remainder has degree at most one, so write it as $ax+b$ and use values at the roots of the divisor. This is a recurring strategy: choose the unknown remainder shape from the degree bound, then determine its coefficients from evaluation data.

For proof exercises such as the roots-of-unity problem, the goal is not to expand a large polynomial. The key is to use the factor theorem twice, once at $\omega$ and once at $\omega^2$, and then exploit $\omega^3=1$. That turns the divisibility of $F(x)+xG(x)$ into two linear equations in the two unknown numbers f(1) and g(1).

Quick checks

Exercises

1. Let $p(x)=3x^4-x^2+2$ and $q(x)=-3x^4+5x+1$. Find the largest possible degree of $p+q$, and 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 Exercise 8.1 from the slides: 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 $x-1$.