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, and any dots that cannot complete a row are 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. The row width is the divisor, the solid rows are the quotient, and the hollow dots are the remainder. Try to break the smallness condition and you cannot: any four hollow dots would have completed another row, so the hollow dots always come out fewer than the row width. That is what “no further copy of the divisor can be pulled out” looks like. Set the count to $12$ against the same rows of $4$ and the hollow dots vanish, the $R = 0$ case where the divisor goes in exactly.

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 which of two polynomials is bigger can flip as $x$ changes. At $x = 2$, $x$ is smaller than $x^2$, since $x^2 = 4$. At $x = \tfrac12$, though, $x$ is larger than $x^2$, since $x^2 = \tfrac14$. 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 has no degree, since it has no leading term, which is why the condition says ”$R = 0$ or $\deg R < \deg D$” instead of just $\deg R < \deg D$. And $D = 0$ is excluded for the same reason $47 \div 0$ is: with $P = 0 \cdot Q + R$, nothing would stop $R$ from just equaling $P$, and the division 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 mirrors a grade-school division almost move for move, with one comfort added: since coefficients never carry, each column settles 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

Two questions are worth answering before you trust the method on harder problems: does every pass keep the identity $P = DQ + R$ true, and does the loop ever actually stop?

Both answers are visible in the two moves you just made. Multiplying the divisor by the new quotient term and subtracting removes exactly that amount from the working remainder and adds it to the quotient. So the total, ”$D$ times the quotient so far, plus what’s left,” never changes: it starts at $D \cdot 0 + P = P$, so it still equals $P$ at every later stage. That is why the final line, whatever it turns out to be, is automatically a true identity.

The quotient term at each pass was chosen for one reason: to cancel the working remainder’s leading term. Once that term is gone, nothing left over can be as large, so the degree of what remains drops by at least $1$ on every pass. A degree cannot drop below $0$, so after a bounded number of passes the working remainder is $0$ or too low in degree to continue, and the loop must stop.

Notice what this quietly explains about the tableau. The invariant is why you can trust the final line without re-deriving anything, and the strict degree drop is why the highest surviving power gets lower after every subtraction. If a row’s highest power does not drop, 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 fix. 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: some quotient and remainder can always be found. A separate question is uniqueness: could a different pair $(Q, R)$ also keep both promises for the same $P$ and $D$? It could not, and seeing why starts from one fact about degrees.

Multiply two nonzero polynomials with leading terms $a x^m$ and $b x^n$. The only way to produce the power $x^{m+n}$ in the product is leading term times leading term, and every other pairing lands strictly lower. Since $a \neq 0$ and $b \neq 0$, their product $ab$ is not zero either, so the product’s leading term really is $ab\,x^{m+n}$:

Degrees add under multiplication. That single fact is enough to rule out any second quotient and remainder. Two different quotients would force $D$ times their difference to have degree at least $\deg D$. But a difference of two too-small remainders can never reach that high, so no second pair exists. However you found a quotient and remainder, by the tableau, by cleverness, or by a lucky guess, they are the quotient and remainder.

Uniqueness pays off immediately in bookkeeping. Suppose $\deg P \geq \deg D$. In $P = DQ + R$ the product $DQ$ has degree $\deg D + \deg Q$. And $R$ is either zero or too low in degree to disturb that leading term, so $P$ and $DQ$ must share a leading term. Two consequences follow, when $\deg P \geq \deg D$,

A cubic divided by a linear divisor must give a quadratic quotient; if it does not, stop and find the error before checking a single coefficient. And if instead $\deg P < \deg D$, 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 $3$.

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

Dividing both sides of $P = DQ + R$ by $D(x)$ gives a second form of the same fact:

This is the polynomial version of turning an improper fraction into a mixed number: $\tfrac{47}{5} = 9 + \tfrac{2}{5}$, quotient out front, remainder still sitting over the divisor, valid wherever $D(x) \neq 0$. From Worked Example 3, for instance,

Both forms carry the same information, but proofs use the identity form. The fraction form is what you’ll reach for later, when a chapter on rational expressions takes up quotients like this one in their own right.

Common mistakes