Quotient and remainder, and how big each one is
Division does not usually come out exactly, and the answer has two parts. (2x³ − 3x² + 4x − 5) ÷ (x − 2) has quotient 2x² + x + 6 and remainder 7, written 2x² + x + 6 + (7) / (x − 2).
Two facts about their sizes are worth holding, because both catch errors:
- The quotient’s degree is the dividend’s minus the divisor’s. A cubic divided by a linear gives a quadratic: 3 − 1 = 2. A quotient of the wrong degree means a stage was skipped.
- The remainder’s degree is strictly below the divisor’s. Dividing by a linear leaves a constant; dividing by (x² − 4) can leave a linear remainder. If your remainder is still as big as the divisor, the division stopped one stage early — and it always goes one stage further than it feels like it should.
When the remainder is zero the division is exact and the divisor is a factor: (x³ − 1) ÷ (x − 1) = x² + x + 1 with nothing left over. That case is the whole content of the factor theorem below.
Long division, and the subtraction that goes wrong
Each stage is four moves: divide the leading term of what is left by the leading term of the divisor, multiply the whole divisor by that result, subtract, and repeat. Nothing about the pattern is unusual — it is the algorithm you already use on numbers.
The hazard is the subtraction. What you subtract is a whole polynomial, so every one of its signs flips, and forgetting that from the second term onwards is the classic failure of the method. Subtracting 2x³ − 4x² means subtracting the 2x³ and adding the 4x². Some people avoid it entirely by negating the product and adding instead, which is the same trick the Adding and Subtracting Polynomials Calculator makes explicit — write the flipped version down as its own line, then add.
The other quiet error is multiplying only the leading term of the divisor rather than all of it. The product at each stage is a full polynomial with as many terms as the divisor has, and a short product leaves a remainder that is too large, which then produces an extra quotient term that should not exist.
Missing powers, and why the zeros matter
3x⁴ − 2x² + 1 has no x³ term and no x term. Written out for division it needs both written in as zeros, because the method is entirely about keeping like terms in the same column, and a gap shifts every column after it.
Dividing it by x + 2 gives 3x³ − 6x² + 10x − 20 with remainder 41. The quotient has all four terms present; nothing about the answer hints that the dividend had gaps, which is exactly why the gaps have to be handled at setup rather than noticed later.
x² + 1 divided by x − 3 is the smallest version of the same thing: no x term in the dividend, and the answer x + 3 + (10) / (x − 3) depends on that column being kept. The calculator inserts the placeholders for you, but on paper they are yours to write.
Synthetic division, and the sentence most sources leave out
Synthetic division drops the variables entirely and works with the coefficients in a row, which makes it fast. It is a genuine shortcut and it has a genuine limit: it only works when the divisor is linear. (x⁴ − 16) ÷ (x² − 4) cannot be done this way at all, and the tool above says so rather than showing an empty table.
Two details cause most of the trouble with it. The first is the sign: dividing by (x − 2) means c = 2, the value that makes the divisor zero, which is the negative of the constant you see written. Dividing by x + 2 means c = −2, not +2.
The second is almost never mentioned. When the divisor is not monic — (2x − 1), say — every coefficient is divided through first, which makes the quotient come out right but leaves the last cell as the remainder of the scaled problem. It has to be multiplied back: −5/2 × 2 = −5. Skipping that step gives a remainder that is wrong by exactly the factor you divided by, and long division on the same problem would disagree with it.
The remainder theorem: skip the division entirely
If you only want the remainder after dividing by (x − c), you do not have to divide at all. The remainder is P(c) — substitute c into the polynomial and that is the answer.
For (2x³ − 3x² + 4x − 5) ÷ (x − 2), substituting 2 gives 7, which is the remainder the long division produced. For 3x⁴ − 2x² + 1 at −2 it is 41, again matching.
The reason is one line of algebra. Writing P(x) = (x − c)·Q(x) + R and putting x = c makes the first term vanish, whatever Q happens to be, leaving P(c) = R. Nothing about Q matters, which is precisely why the shortcut works — and why it applies only to a linear divisor, since only then is the remainder a constant.
The factor theorem: a root and a factor are the same fact
Take the remainder theorem and set the remainder to zero. P(c) = 0 exactly when (x − c) divides P with nothing left over — so c is a root if and only if (x − c) is a factor. Two ideas that are taught separately turn out to be one idea seen from two sides.
P(1) = 0, so by the factor theorem x − 1 divides x³ − 1 exactly, and 1 is a root.
P(2) = 7, which is not zero, so x − 2 is not a factor and 2 is not a root.
This is what makes division useful rather than merely possible. Find one root of a cubic by any means — the rational root candidates are usually the fastest — divide it out, and you are left with a quadratic that the quadratic formula finishes. That pairing is how most higher-degree polynomials actually get solved by hand, and division is the step in the middle.
Sources and methodology
The division algorithm and both theorems are proved in a line each above rather than asserted — the remainder theorem in particular is short enough that citing it would be less convincing than showing it. What is cited is the standard reference statement.
Method. Every figure on this page comes from src/lib/polynomial-division.ts over the exact-rational polynomial core in src/lib/algebra/poly.ts. No answer is asserted: the engine multiplies the divisor by the quotient, adds the remainder, and compares the result with the dividend coefficient by coefficient before anything is displayed. The synthetic table’s scaled remainder is reconciled with the long-division remainder rather than left as the raw bottom-row value, which is the detail most implementations get wrong on a non-monic divisor. That engine is verified on every change against 77 hand-written assertions, including that the two methods agree on every linear-divisor case swept and that the remainder always equals P(c). The count and the per-case breakdown are published on the formula verification page.
Read the guide
Division is the last of four operations and the one that depends most on the other three being solid. Adding, Multiplying and Dividing Polynomials — the Four Operations puts them together: the sign flip that subtraction needs and that appears again inside every stage of long division, the product count that catches a dropped term, the placeholder zeros, and the two checks that work on all four.