Polynomial Division

Just as integers can be divided to produce a quotient and a remainder, polynomials can be divided by other polynomials. The structure is identical: divide, multiply, subtract, and repeat until the leftover piece is too small to divide further. This section establishes the formal framework before we study the algorithms.

Quick Reference

Term	Meaning
Dividend $f (x)$	The polynomial being divided
Divisor $d (x)$	The polynomial we divide by
Quotient $q (x)$	Result of the division
Remainder $r (x)$	What is left over; $\deg (r) < \deg (d)$
Division algorithm	$f (x) = q (x) d (x) + r (x)$
Degree relationship	$\deg (q) = \deg (f) - \deg (d)$
Divisibility	$f$ is divisible by $d$ when $r (x) = 0$

The Division Algorithm

Let $f (x)$ and $d (x)$ be polynomials with $d (x) \neq 0$ . There exist unique polynomials $q (x)$ (the quotient) and $r (x)$ (the remainder) such that

f (x) = q (x) d (x) + r (x),

where either $r (x) = 0$ or $\deg (r) < \deg (d)$ .

We say $f (x)$ is divisible by $d (x)$ if the remainder $r (x) = 0$ .

The condition $\deg (r) < \deg (d)$ is essential: it means the remainder cannot be divided further by the divisor without introducing fractional terms. This is the exact analogue of the fact that when dividing integers, the remainder is always smaller than the divisor.

Degree of the Quotient

If $\deg (f) = n$ and $\deg (d) = m$ (with $m \leq n$ ), then $\deg (q) = n - m$ .

Why the degree of the quotient is

n - m

For any two nonzero polynomials

P

and

Q

, the degree of their product is the sum of their degrees:

\deg (P Q) = \deg (P) + \deg (Q) .

In the division equation

f (x) = q (x) d (x) + r (x)

, the remainder has degree less than

m

, so it cannot contribute to the leading term of

f

. Therefore the leading term of

f

must come entirely from

q (x) d (x)

\deg (f) = \deg (q d) = \deg (q) + \deg (d) .

Substituting

\deg (f) = n

and

\deg (d) = m

gives

\deg (q) = n - m

Uniqueness of the Quotient and Remainder

For given polynomials $f (x)$ and $d (x) \neq 0$ , the quotient $q (x)$ and remainder $r (x)$ satisfying the division algorithm are unique.

Proof

Suppose two pairs

(q_{1}, r_{1})

and

(q_{2}, r_{2})

both satisfy the division algorithm:

f (x) = q_{1} (x) d (x) + r_{1} (x) and f (x) = q_{2} (x) d (x) + r_{2} (x) .

Subtracting gives:

(q_{1} (x) - q_{2} (x)) d (x) = r_{2} (x) - r_{1} (x) .

Both

r_{1}

and

r_{2}

have degree less than

\deg (d)

, so the right side has degree less than

\deg (d)

. If

q_{1} \neq q_{2}

, the left side would have degree at least

\deg (d)

, a contradiction. Therefore

q_{1} = q_{2}

, and consequently

r_{1} = r_{2}

How to Divide

What remains is to show how f(x) can be expressed in the form q(x)d(x) + r(x), where r(x) has a degree lower than d(x) or is equal to zero. The most common method for achieving this is called the "long division" — a process analogous to the long division of numbers — which is illustrated in the following example.

Frequently Asked Questions

Why must the remainder have degree less than the divisor?

If the remainder had degree greater than or equal to the divisor, we could continue dividing: we could extract at least one more term for the quotient. We stop exactly when the remaining piece is "too small" to divide, which means its degree is strictly less than the degree of the divisor.

What happens when the dividend has a lower degree than the divisor?

\deg (f) < \deg (d)

, then

f

itself is the remainder and the quotient is 0. For example, dividing

x + 1

x^{2} + 1

gives quotient 0 and remainder

x + 1

, because

x + 1

cannot be reduced further by a degree-2 divisor.

How is polynomial division related to factoring?

If the remainder is 0, then

f (x) = q (x) \cdot d (x)

, meaning

d (x)

is a factor of

f (x)

. Division is the primary tool for testing and confirming factorizations.