Polynomial Long Division

The promise integer division makes

Before touching a polynomial, look closely at what integer division actually delivers, because the polynomial version copies it clause for clause.

Divide $47$ by $5$. The process returns $9$ with $2$ left over, and the entire content of that answer is one equation together with one condition:

The equation says the answer is correct: nine fives and two more really do rebuild $47$. The condition says the division is finished: fewer than five remain, so no further five can be pulled out. Drop the condition and “answers” multiply without end. The equations $47 = 5 \cdot 8 + 7$ and $47 = 5 \cdot 6 + 17$ are both true, yet nobody calls $8$ remainder $7$ the result of dividing $47$ by $5$, because a remainder of $7$ still contains a five. Only the demand that the remainder be smaller than the divisor pins down the pair $(9, 2)$ as the one and only answer.

So division with remainder is really two promises: an identity, dividend equals divisor times quotient plus remainder, and a smallness condition that makes the identity’s pieces unique. Keep both promises in view; the whole lesson is about keeping them for polynomials.

Both promises are visible at once if you lay the dividend out in rows. In the figure below you set a number of dots and a row width. The dots fill rows from the top left, with any that cannot complete a row drawn hollow.

Start at $11$ dots in rows of $4$. Two rows fill and three dots are left hollow, which is the identity $11 = 4 \cdot 2 + 3$ drawn rather than computed. In that picture the row width is the divisor, the solid rows are the quotient, and the hollow dots are the remainder. Now try to break the smallness condition. You cannot, and the reason is worth seeing rather than being told: any four hollow dots would have completed a row and stopped being hollow. Whatever count and width you set, the hollow dots come out fewer than the row width. That is the same statement as “the rows are as full as they go”. That in turn is the same statement as “no further copy of the divisor can be pulled out”. Set the count to $12$ against the same rows of $4$ and the hollow dots vanish altogether, the $R = 0$ case where the divisor goes in exactly. That is the entire promise of division with remainder. The rest of this lesson is the work of making both halves of that promise survive the move from counting dots to counting degrees.

A polynomial is a numeral written in base x

The last lesson gave you the vocabulary. The degree of a polynomial is its highest power, the leading term carries that power, and standard form lists the terms in falling powers. Now set a numeral and a polynomial side by side:

Same skeleton. The coefficients play the digits, and the powers of $x$ play the places. Two honest differences separate them. First, a numeral’s digits stay between $0$ and $9$, and any overflow is handled by carrying into the next column. A polynomial’s coefficients, by contrast, are unrestricted, so nothing ever carries and the columns never interact. In that one respect polynomial arithmetic is simpler than integer arithmetic. Second, a numeral names a single number, while a polynomial’s value moves with $x$. That second difference is why the smallness condition needs translating. The phrase “the remainder is smaller than the divisor” cannot mean numerical size, because $x - 100$ is smaller than $x^2$ for some inputs and larger for others. The measure that does not move with $x$ is the degree, so degree takes over the job that size did for integers.

With that translation made, here is the exact claim this lesson runs on, proves, and then uses. It is called the division algorithm for polynomials. Let $P(x)$ be any polynomial (the dividend) and $D(x)$ any nonzero polynomial (the divisor). Then there is exactly one pair of polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) with

Two small print items. The zero polynomial is not assigned a degree, since it has no leading term, which is why the condition names it separately instead of writing $\deg R < \deg D$ alone. And the divisor $D = 0$ is excluded for the same reason $47 \div 0$ is. The equation $P = 0 \cdot Q + R$ would force $R = P$ with no smallness condition left to satisfy, and division by zero would promise nothing.

Each quotient term is forced

The algorithm is a loop, and the loop has only one legal move. Watch it once in slow motion on

Your goal is to peel copies of $x + 2$ out of $x^2 + 5x + 7$ until what is left is too small, in degree, to contain another copy. The tool is subtraction: subtract a multiple of the divisor, record how many copies that multiple used, repeat. To make progress you must kill the dividend’s leading term $x^2$, and here is the key point: a multiple $q \cdot (x + 2)$ cancels $x^2$ exactly when its own leading term is $x^2$. If you take $q$ to be a single term, the leading term of $q \cdot (x + 2)$ is just $q \cdot x$, so you need $q \cdot x = x^2$, which forces

Divide the leading terms: that is the whole rule, and no other choice touches the $x^2$. Multiply back and subtract, remembering to subtract the entire product:

The $x^2$ is gone, exactly as designed. What remains is a smaller division problem, $3x + 7$ divided by $x + 2$, so run the same forced move again: $3x \div x = 3$, then $3(x + 2) = 3x + 6$, and $(3x + 7) - (3x + 6) = 1$. Now the working remainder $1$ has degree $0$, which is below the divisor’s degree $1$. The loop must stop, because any further multiple of $x + 2$ would introduce a new $x$ term rather than cancel anything. Collecting the two recorded quotient terms, $x$ and $3$,

The traditional tableau records exactly these moves in columns, quotient on the roof, each product written under the matching powers:

Now cash in the base-$x$ idea. Substitute $x = 10$ everywhere: the dividend becomes $157$, the divisor becomes $12$, the quotient becomes $13$, and the remainder stays $1$. Check it as an integer sentence: $12 \cdot 13 + 1 = 156 + 1 = 157$. The division you just performed is a grade-school division in disguise, column for column, with one comfort added: since coefficients never carry, each column settled in a single step.

One habit before moving on. Every polynomial division can be checked by multiplying back, and the check is cheaper than the division was:

Make the check a reflex. The division algorithm’s identity is an equation about every value of $x$ at once, so expanding $DQ + R$ and comparing with $P$ catches any slip immediately.

Why the loop is correct, and why it must stop

The demonstration above worked, but a method you will use for the rest of the course deserves more than one lucky run. Two things need proving: the loop never breaks the identity it is building, and the loop cannot run forever.

Notice what the proof quietly explains about the tableau. The invariant is why you may trust the final line without re-deriving anything. The strict degree drop is why each row of the tableau sits at least one column further right than the row above it. When a row fails to move right, a subtraction error has slipped in, usually a sign.

Missing terms need placeholders

The tableau is column bookkeeping, and columns only work when every power has a column. A numeral handles a skipped place with the digit $0$. The zero in $3{,}205$ contributes nothing to the sum, but it holds the tens column open so the $2$ and the $3$ land where they belong. A polynomial that skips powers needs the same courtesy. Before dividing, rewrite the dividend (and the divisor, if need be) with an explicit $0$ coefficient for every missing power. Nothing changes mathematically, since $0 \cdot x^k$ is zero; what changes is that like powers now sit above one another in the tableau, where the subtractions can find them.

Only one answer is possible

Existence came from running the loop. Uniqueness, the claim that no different pair $(Q, R)$ could also satisfy both promises, needs one degree fact first. Suppose $f$ and $g$ are nonzero polynomials with leading terms $a x^m$ and $b x^n$. Multiplying them, the only way to produce the power $x^{m+n}$ is leading term times leading term, and every other pairing lands strictly lower. The coefficient produced there is $ab$, and $ab \neq 0$ because $a \neq 0$ and $b \neq 0$. A product of two nonzero real numbers is never zero, which is the Zero Product Property from the quadratics chapter read in reverse. So the product’s leading term is $ab\,x^{m+n}$, and

Degrees add under multiplication. That single fact powers the uniqueness argument.

Uniqueness pays an immediate dividend in bookkeeping. Suppose $\deg P \geq \deg D$, so the quotient $Q$ is not zero. In $P = DQ + R$ the product $DQ$ has degree $\deg D + \deg Q$, and $R$ is either zero or too low in degree to interfere with that leading term. So the leading terms of $P$ and $DQ$ must be identical. Two consequences: the degrees satisfy

and the leading coefficient of $Q$ is the dividend’s leading coefficient divided by the divisor’s. A cubic divided by a linear divisor must give a quadratic quotient; if it does not, stop and find the error. Use both facts as a pre-flight check before you divide and as a sanity check after.

One more consequence: if $\deg P < \deg D$, the pair $Q = 0$, $R = P$ already keeps both promises, so by uniqueness it is the answer. Dividing $x + 1$ by $x^2 + 1$ gives quotient $0$ and remainder $x + 1$, just as $3 \div 5$ gives quotient $0$ and remainder $3$. The loop agrees: its stopping condition is met before the first pass.

A divisor of higher degree

Nothing in the loop, the proof, or the bookkeeping assumed the divisor was linear. Divide by a quadratic and the only visible change is the finish line. With that divisor the loop stops when the working remainder is zero or of degree below $2$, so a remainder may now be linear, or a constant, or zero.

Two ways to package the answer

The identity $P = DQ + R$ has a twin obtained by dividing both sides by $D(x)$:

This is the polynomial version of converting an improper fraction to a mixed number: $\tfrac{47}{5} = 9 + \tfrac{2}{5}$, quotient out front, remainder still sitting over the divisor. From Worked Example 3, for instance,

Both packagings carry the same information, but they are not interchangeable. The identity form holds for every value of $x$ with no exclusions, which makes it the right form for proofs, like the ones in this lesson. The fraction form must dodge the divisor’s zeros. That form becomes the workhorse later in the course, when the chapter on rational expressions takes up quotients like these in their own right.

Common mistakes