Synthetic Division Calculator

Write the root of the divisor in a box, then list the dividend's coefficients in a row, using 0 for any missing power. Bring the first coefficient down, multiply it by the box number, add the result to the next coefficient, and repeat. The last number is the remainder and the others are the quotient.

Think of it as long division with the variables removed. You only keep the coefficients, flip the divisor's sign once at the start, and then the same two moves repeat: multiply the last result by that number and add. Because subtraction turns into addition, there are fewer chances for a sign mistake.

For a divisor like x − 3 it is usually easier than long division: it needs only multiplication and addition and takes a few lines. The usual trouble spots are the sign of the number in the box and forgetting a zero placeholder. With those two habits it is a reliable method.

Synthetic division is easier when the divisor is linear, such as x − c. It uses less writing and avoids subtraction. Long division is easier when the divisor has degree two or higher, because synthetic division cannot handle those at all. Choose by the divisor, not by preference.

Set the divisor equal to zero and solve for x. For x − 4 the number is 4, for x + 4 it is -4, and for 2x − 1 it is 1/2. That value is the root of the divisor, and it goes in the box at the left of the table. The sign of the constant term flips.

Divide 2x³ − 6x² + 2x − 1 by x − 3. Put 3 in the box and write 2, -6, 2, -1. Bring down 2, multiply by 3 to get 6, add to -6 for 0, then 0 × 3 = 0 added to 2 gives 2, and 2 × 3 = 6 added to -1 gives 5. The quotient is 2x² + 2 with remainder 5.

Yes, the calculator above is one. Enter the dividend and the divisor, or enter the root directly, and it draws the synthetic division table, lists each multiply-and-add move, reads off the quotient and remainder, and checks the result with the remainder theorem. Fractions and decimals are handled exactly.

If you only know a root c, use the root mode: enter the dividend and c, and the calculator divides by x − c. This is the situation when you are testing whether a number is a zero of the polynomial. A remainder of 0 confirms c is a root and x - c is a factor.

The method works the same way with a complex number in the box, but this calculator accepts real numbers and fractions only. To divide by x − (2 + i) by hand, multiply and add exactly as usual, using complex arithmetic for each multiplication. Conjugate pairs of roots are often easier to handle by dividing by a quadratic.

There is no single closed formula, but the rule is short. With coefficients a_n down to a_0 and divisor x − c, set b_(n-1) = a_n, then b_(k-1) = a_k + c x b_k, and the remainder is a_0 + c x b_0. Each new number comes from the previous one times c, plus the next coefficient.

It is polynomial long division with the variables, repeated terms and subtractions stripped out. Each column performs one round of long division: multiply the divisor by the new quotient term and subtract. Using the root with its sign flipped turns that subtraction into the addition you see in the table.

Long division subtracts the divisor times each quotient term. Dividing by x - c means subtracting c times the quotient term from the next coefficient, and flipping the sign to use c lets you add instead. The flip also puts the root of the divisor in the box.

Write a zero for every power that does not appear. For x⁴ − 16 the coefficients are 1, 0, 0, 0, -16. Skipping a zero shifts every later column and gives a wrong quotient. The calculator inserts the placeholders for you and says which ones it added.

It means the divisor divides the polynomial exactly. The divisor is a factor, the number in the box is a root, and the dividend equals the divisor times the quotient. This is the factor theorem in action and the basis of finding all the roots of a cubic or quartic step by step.

The remainder theorem says that dividing a polynomial p(x) by x - c leaves a remainder equal to p(c). So the last number in a synthetic division table is the value of the polynomial at c. It is a quick way to evaluate a polynomial, needing roughly half the multiplications of plugging in.

The factor theorem says x - c is a factor of p(x) exactly when p(c) = 0, which is exactly when the synthetic division remainder is 0. Once you find one root c, the quotient is a lower-degree polynomial you can factor or solve further, which is how cubics get broken into linear and quadratic pieces.

List the possible rational roots using the rational root theorem, test each with a synthetic division table, and keep any that give remainder 0. Each success lowers the degree by one. Repeat on the quotient until a quadratic is left, then solve it directly or with the quadratic formula.

For a polynomial with integer coefficients, any rational root p/q in lowest terms has p dividing the constant term and q dividing the leading coefficient. For 2x³ − 3x² − 11x + 6 the candidates come from 1, 2, 3, 6 over 1, 2. It narrows an endless search to a short list to test.

Use c = -b/a in the box, run the table as usual, then divide every quotient coefficient by a. The remainder stays the same. For 4x³ − 3x + 7 divided by 2x − 1, use c = 1/2, get 4, 2, -2 with remainder 6, then halve to get 2x² + x − 1.

No. Classic synthetic division needs a divisor of degree one, like x − 3 or 2x + 1. Divisors with x² or higher need polynomial long division. The same applies to dividing by x² − 4 directly, although you can factor it into (x − 2)(x + 2) and divide by each factor in turn.

The quotient has degree one less than the dividend. A cubic divided by a linear factor gives a quadratic, and a quartic gives a cubic. The remainder is always a plain number. If the dividend is a constant, the quotient is 0 and the remainder is that constant.

Write the quotient, then add the remainder over the divisor. For (2x³ − 6x² + 2x − 1) ÷ (x − 3) the answer is 2x² + 2 + 5/(x − 3). If the remainder is 0, the answer is just the quotient, and you can say the divisor is a factor.

Multiply the quotient by the divisor and add the remainder. You should get the original dividend back. You can also check the remainder by evaluating the polynomial at c, which must match. The calculator performs the second check automatically and reports the value of p(c).

Yes. Fractions in the box and in the coefficients work the same way, they just need careful arithmetic by hand. This calculator uses exact fractions, so a root of 1/2 or a coefficient of 2/3 gives exact results with no rounding. Decimals are converted to exact fractions first.

Dividing by x means c = 0. The table copies each coefficient down unchanged, so the quotient is the dividend with its constant term removed and shifted down a power, and the remainder is that constant. It is a quick way to see that p(0) is the constant term.

The special case of dividing by x - r is known as Ruffini's rule after Paolo Ruffini, who described it in 1809. A closely related evaluation scheme is Horner's method. Synthetic division generalizes the idea, though most courses teach only the linear monic case.

They use the same multiply-and-add rhythm. Horner's method evaluates a polynomial at a point, and the synthetic division table does the same job while also recording the quotient. The remainder of the table is exactly Horner's value p(c), so many people treat them as two views of one calculation.

Each column handles one power. After the constant term column nothing is left to divide, so the final sum is whatever the divisor cannot absorb, which is a number with no x in it. That makes it the remainder, and by the remainder theorem it equals p(c).

It is the other factor. If p(x) divided by x - c leaves 0, then p(x) = (x − c) times the quotient. For x³ − 6x² + 11x − 6 divided by x − 1, the quotient x² − 5x + 6 factors as (x − 2)(x − 3), so p(x) = (x − 1)(x − 2)(x − 3).

Yes, repeatedly. After dividing by x − c, run the table again on the quotient with the same c to test for a double root. It is also how you find all the roots of a high-degree polynomial: divide, reduce the degree, and test the next candidate on the smaller quotient.

Yes. Graph the polynomial and look where it crosses the x-axis: those crossings are the real roots, and a crossing near a whole number or simple fraction is a good candidate for the box. Confirm it with synthetic division, because a graph only suggests a root while a zero remainder proves it.

Yes. If the divisor is x + 3, the root is -3, and you multiply by -3 at each step. Negative numbers in the box make the middle row alternate in sign, so take care with each product. The calculator prints each multiplication so you can compare your own work.

Synthetic division is a method that produces the whole quotient and the remainder. The remainder theorem is a fact: the remainder equals p(c). Use the theorem when you only need p(c), and the method when you want the quotient as well. The last number of the table gives both.

A dividend of degree n has n + 1 coefficients, so the table has n + 1 columns and needs n multiplications and n additions. A cubic needs three of each. That is why it is faster than long division, which repeats a full multiply-and-subtract of the divisor for every quotient term.

This page is one. It shows the number to divide by, the coefficient row with zero placeholders, each multiply-and-add move, the finished table, the quotient and remainder, and a remainder theorem check. It also tells you whether the divisor is a factor of the dividend.