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: $\deg(PQ) = \deg(P) + \deg(Q).$ 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: $f(x) = q_1(x)\,d(x) + r_1(x) \qquad \text{and} \qquad f(x) = q_2(x)\,d(x) + r_2(x).$ 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.
