Synthetic Division Calculator

A synthetic division calculator divides a polynomial by a linear factor such as (x − 2) using only the coefficients, so the whole problem fits in three short rows of numbers. Divide 2x⁴ − 3x³ − 5x² + 3x + 8 by (x − 2) and the quotient is 2x³ + x² − 3x − 3 with a remainder of 2, in eight arithmetic steps.

Three sister tools cover the neighboring questions. The Polynomial Calculator adds, subtracts and multiplies polynomials but does not divide them, so use it to build or expand an expression before you bring it here. The Long Division Calculator handles plain numbers rather than polynomials, and the Quadratic Formula Calculator takes over once synthetic division has reduced your polynomial to a quadratic and you need its roots. Use this page whenever the divisor is linear and you want the quotient, the remainder and the working shown side by side.

What synthetic division actually is

Every division of one polynomial by another can be written in the same shape as ordinary whole-number division: a dividend, a divisor, a quotient and a remainder. When the divisor is linear and its leading coefficient is 1, almost everything you would write out in polynomial long division is bookkeeping. The x symbols, the exponents and the alignment of like terms never change; only the coefficients do any work. Synthetic division strips the notation away and keeps just those coefficients, which is why it takes a fraction of the space and far less time.

The trade-off is that it is narrow. Synthetic division is defined for a divisor of the form (x − c). Anything else, a quadratic divisor for instance, needs full polynomial long division or a factoring workaround. Within its lane, though, it is not merely a shortcut but the fastest known way to both divide by a linear factor and evaluate a polynomial at a point, which is why the same arithmetic pattern appears in numerical software under the name Horner evaluation.

The formula behind the table

Write the polynomial with coefficients a_n down to a₀ and divide by (x − c). The quotient coefficients b are produced by a single recurrence:

Applied from the leading coefficient across to the constant, this produces one number per column. The last value it generates is not a quotient coefficient but the remainder R, and the whole calculation satisfies the division identity:

That second equality is the remainder theorem, and it makes synthetic division double as an evaluation tool: the bottom-right number is the value of the polynomial at x = c. When it comes out as 0 you have found a root, and the identity collapses to P(x) = (x − c) × Q(x), which is the factor theorem.

Worked example: dividing a quartic by (x − 2)

Take P(x) = 2x⁴ − 3x³ − 5x² + 3x + 8 and divide it by (x − 2). The divisor is zero at x = 2, so the number in the box is 2. The coefficients, in descending order, are 2, −3, −5, 3 and 8; no power is missing, so no zero placeholders are needed. The work runs like this:

Reading the steps out loud: bring down 2. Multiply 2 × 2 = 4, write it under −3 and add to get 1. Multiply 1 × 2 = 2, write it under −5 and add to get −3. Multiply −3 × 2 = −6, add to 3 and get −3. Multiply −3 × 2 = −6, add to 8 and get 2. That final 2 is the remainder.

The bottom row therefore reads 2, 1, −3, −3 for the quotient and 2 for the remainder. Because the dividend had degree 4, the quotient has degree 3:

You can verify the whole thing two ways. Expand the right-hand side and you get the original polynomial back exactly. Or evaluate P(2) directly: 2(16) − 3(8) − 5(4) + 3(2) + 8 = 32 − 24 − 20 + 6 + 8 = 2, matching the remainder, exactly as the remainder theorem promises.

Second example: when the remainder is zero

Now divide P(x) = x³ − 4x² + x + 6 by (x + 1). The divisor is written with a plus sign, so the number in the box is −1, not 1. Bring down 1. Multiply 1 × (−1) = −1 and add to −4 to get −5. Multiply −5 × (−1) = 5 and add to 1 to get 6. Multiply 6 × (−1) = −6 and add to 6 to get 0.

The remainder is 0, so (x + 1) is a factor and x = −1 is a root. The quotient is x² − 5x + 6, which factors by inspection into (x − 2)(x − 3). Put it together and the cubic is completely factored:

The roots are −1, 2 and 3. This two-step pattern, peel off one linear factor with synthetic division and then solve the smaller polynomial that remains, is how most cubic and quartic problems in an algebra course are meant to be solved. If the leftover piece is a quadratic that does not factor nicely, hand it to the Quadratic Formula Calculator.

Which number goes in the box

The box always takes the value of x that makes the divisor equal to zero, which flips the sign you see written. This is the step that trips people up more than any other, so the table below spells it out for the divisor shapes you are likely to meet.

The last two rows deserve a sentence of their own. A divisor like 2x − 5 factors as 2(x − 2.5), so you run the synthetic step with c = 2.5 and then scale the quotient down by the 2 you factored out. The remainder is already correct and must not be divided. Dividing 2x³ − 3x² − 11x + 6 by (2x − 5) works out to the raw bottom row 2, 2, −6 with remainder −9, and halving the quotient gives x² + x − 3, remainder −9.

How to use this calculator

1. Pick the degree. Choose the highest power that appears in your polynomial, from 2 up to 8. The number of coefficient boxes updates to match, always one more than the degree.
2. Type the coefficients. Fill them in from the highest power on the left down to the constant on the right. Every box is labeled with its power, and any power your polynomial does not use simply gets a 0.
3. Choose the divisor form. Tap x − c or x + c to match how your problem is written, then enter c as a positive number. The line underneath confirms the divisor and shows the root that will actually be used, so a sign slip is caught before you read the answer.
4. Read the quotient. The large result at the top is the quotient in standard form, with the remainder shown beside it and the factor-theorem verdict in the tiles below.
5. Check the working. The synthetic division table reproduces the three-row layout you would write by hand, the numbered step list narrates every multiply-and-add, and the division identity restates the answer as a single equation you can expand to verify.

Everything recalculates as you type, so you can run a whole list of candidate roots through the same polynomial in a few seconds simply by editing the c value.

Who this calculator is for

• Algebra 2 and pre-calculus students checking homework where the assignment is explicitly to divide with the synthetic method.
• Anyone hunting for roots of a cubic or quartic, testing candidates from the rational root theorem one at a time.
• Students verifying a factorization, since a remainder of 0 is the fastest proof that a proposed linear factor really divides the polynomial.
• Calculus students simplifying a rational function before integrating or before finding a slant asymptote, both of which need the quotient-plus-remainder form.
• Teachers and tutors who need worked examples with the full table already laid out.
• Test takers who want to evaluate a polynomial at a point quickly, because the remainder is exactly that value.

Ten worked divisions to check against

Each row below was produced with the same synthetic division routine that powers the calculator. Reproduce any of them by entering the coefficients and the divisor.

Two patterns are worth noticing. Rows with a remainder of 0 are exactly the ones where the divisor is a genuine factor, and in the x⁴ − 16 row the missing x³, x² and x terms had to be entered as three zeros before the division would work.

Using the remainder theorem to find roots

The rational root theorem narrows the candidates: any rational root of a polynomial with integer coefficients is a factor of the constant term divided by a factor of the leading coefficient. For x³ − 4x² + x + 6 the leading coefficient is 1 and the constant is 6, so the candidates are the divisors of 6 with both signs. Running each through synthetic division gives the remainder, which by the remainder theorem is the value of the polynomial there.

Three candidates return 0, and a cubic has at most three roots, so the search is finished: the roots are −1, 2 and 3, matching the factorization found earlier. In practice you stop at the first success, because the quotient it leaves is a quadratic you can solve outright instead of testing the remaining candidates.

Synthetic division compared with long division

Both methods answer the same question and always agree. What differs is the effort. Dividing a degree-n polynomial by (x − c) with synthetic division costs exactly n multiplications and n additions: 6 operations for a cubic, 8 for a quartic, 16 for a degree-8 polynomial. Long division performs the same arithmetic but wraps it in written subtractions of full expressions, and every one of those subtractions is a chance to drop a sign.

Long division earns its keep in the cases synthetic division cannot touch. If the divisor is quadratic or higher, if it is irreducible like x² + 1, or if you need the intermediate partial remainders written out as polynomials, long division is the tool. A useful middle path exists for divisors that factor: to divide by x² − 4, run synthetic division by (x − 2) and then divide that quotient by (x + 2). Two quick passes replace one slow one.

Key terms

• Dividend: the polynomial being divided, written P(x) here.
• Divisor: what you divide by. Synthetic division requires the linear form (x − c).
• Quotient: the result of the division, always one degree lower than the dividend.
• Remainder: what is left over, a single number for a linear divisor, and equal to P(c).
• Leading coefficient: the number attached to the highest power, the first value you bring down.
• Standard form: terms written from the highest power down to the constant, which is the order the coefficient boxes expect.
• Root or zero: a value of x that makes the polynomial 0, which shows up here as a remainder of 0.
• Horner evaluation: the nested multiply-and-add scheme used to compute P(c); it is arithmetically identical to synthetic division.

What changes the result

• The sign of c: the largest single source of wrong answers. Dividing by (x + 4) uses −4, and using +4 instead produces a completely different quotient and remainder.
• Missing powers: forgetting a zero placeholder shifts every later coefficient into the wrong column and corrupts the whole bottom row.
• Coefficient order: the boxes read highest power first, so a polynomial typed in ascending order divides something else entirely.
• The leading coefficient of the divisor: anything other than 1 requires factoring it out and scaling the quotient afterwards.
• Decimal precision: exact fractions such as one third can only be entered as a rounded decimal, which leaves a tiny remainder where an exact calculation would give 0.

Tips for getting it right first time

• Rewrite the polynomial in standard form before you read off a single coefficient.
• Count the boxes: a degree-n polynomial has n + 1 coefficients, so if the count is off, a term is missing.
• Test c = 1 first when hunting for roots; the remainder is just the sum of all the coefficients, which you can often do in your head.
• Expand your answer, (x − c) times the quotient plus the remainder, whenever the stakes are high. It should rebuild the original polynomial exactly.
• After a remainder of 0, keep dividing the quotient by the next candidate root rather than starting over from the original polynomial. Each pass is shorter than the last.
• Watch for a leading coefficient other than 1 in the dividend, which is fine, versus one in the divisor, which is not.

Limitations and assumptions

• The divisor must be linear. Quadratic and higher divisors need polynomial long division.
• Coefficients and c are treated as real decimal numbers. Exact fractions and complex numbers are outside its scope, so a root such as one third is approximated.
• Degrees from 2 through 8 are supported, which covers essentially all classroom work.
• It divides but does not factor. It confirms whether a candidate factor works, and leaves choosing the candidates to you or to the rational root theorem.
• Results are rounded to remove binary floating-point noise, so an extremely small non-zero remainder from a rounded decimal input is displayed as 0.

Which calculator to use when

Use this page when the divisor is linear and you want the quotient, the remainder and the working. Use the Polynomial Calculator to add, subtract or multiply polynomials, which is also the fastest way to expand (x − c) times your quotient and confirm the answer. Use the Quadratic Formula Calculator once synthetic division has reduced the problem to a quadratic. Use the Factoring Calculator to find the factors of the constant term when you are listing rational root candidates, and the Long Division Calculator when the numbers themselves, not polynomials, are what you need divided.

Sources

This calculator relies on no external data. Synthetic division, the remainder theorem and the factor theorem are exact results of polynomial algebra, and every number on this page, in the tables and in the worked examples, follows deterministically from the coefficients and the divisor you enter. Specifically:

• The recurrence b_k−1 = a_k + c × b_k is the definition of synthetic division and needs no source; it is verified here by expanding (x − c) times the quotient plus the remainder.
• The remainder theorem, R = P(c), and the factor theorem, R = 0 exactly when (x − c) divides P(x), are standard theorems of algebra proved from the division identity itself.
• The remainder shown by this tool is cross-checked against P(c) computed independently by Horner evaluation, so the two agree for every input.
• No physical units, financial figures, tax values or measured constants are used anywhere on this page, so no primary data source applies.