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Synthetic Division Calculator

Created by Anna Szczepanek, PhD
Reviewed by Dominik Czernia, PhD and Jack Bowater
Last updated: May 21, 2024
Table of contents:

Welcome to our synthetic division calculator! It helps you perform the synthetic division of polynomials while showing all the intermediate steps at the same time!

Have you ever wondered what synthetic division is? Do you need to learn how to do synthetic division? We teach you everything you need to know about dividing polynomials using synthetic division, provide examples of synthetic division with steps, and explain how to use synthetic division to find zeros.

As a bonus, we show you how to deal with non-monic and quadratic divisors!

Synthetic division of polynomials – definition

Before we can explain how to divide polynomials using synthetic division, let's refresh a few basic notions:

What are polynomials?

A polynomial is an expression involving a sum of non-negative integer powers of at least one variable, each multiplied by real (or complex) numbers, which we call coefficients.

A polynomial in one variable, x (a univariate polynomial), is given by

anxn + an-1xn-1 + ... + a1x + a0,

where an, an-1,..., a1, a0 are the coefficients. We call the individual terms of the form akxk monomials. The leading coefficient of this polynomial is the coefficient of the term with the highest power of x, i.e., the coefficient an, provided that an ≠ 0. We say a polynomial is monic if its leading coefficient is equal to one: an = 1.

The degree of a polynomial is the value of the greatest exponent present in the polynomial with a non-zero coefficient. The polynomial written above has degree n, provided that an ≠ 0. Constant non-null polynomials have degree zero. A null polynomial has its degree left undefined or, sometimes, defined as -∞ (negative infinity). We usually denote the degree of a polynomial with deg.

Polynomial division

The division of polynomials is analogous to dividing integers with remainder, which you've most probably encountered in arithmetic. Let P(x) and D(x) be two polynomials. If D(x) is non-zero, then there exist two polynomials, Q(x) and R(x), which satisfy:

P(x) = D(x) ⋅ Q(x) + R(x)

and deg(R) < deg(D). Moreover, Q(x) and R(x) are unique, i.e., there's no other pair of polynomials that satisfy these two conditions.

The terms we use in polynomial division are analogous to those in arithmetic: P(x) is called the dividend, D(x) is the divisor, Q(x) is a quotient, and R(x) is the remainder.

Note that:

  • R(x) = 0 if, and only if, P(x) has D(x) as a factor; and
  • If deg(P) < deg(Q), then D(x) = 0 and P(x) = R(x).

The standard way of calculating the quotient and remainder, given a dividend and divisor, is via the algorithm called the polynomial long division.

What is the synthetic division of polynomials?

Synthetic division is a shortcut way of dividing polynomials. It gives the same results as the polynomial long division but is much faster as it involves only the coefficients of the dividend and divisor, on which we perform basic arithmetic operations. As a result, we obtain the coefficients of the quotient and the remainder.

At a first look, you may find synthetic division a bit complicated, but rest assured: once you get the hang of it, you'll never look back!

Synthetic division is most commonly used when dividing by linear monic polynomials x - b. Dividing by such polynomials is very important in the context of finding zeroes and factoring polynomials: to verify whether b is a root of a polynomial, we can synthetically divide this polynomial by x - b and check if the remainder is equal to zero. For details, check out the section below, where we discuss how to use synthetic division to find the zeros of a polynomial.

Keep in mind that synthetic division works for any polynomial divisors: for non-monic polynomials as well as for polynomials of degrees higher than one. However, it becomes more and more complicated as the degree of the divisor grows. In this article, we'll discuss in detail some synthetic division examples of non-monic linear polynomials b1x + b0 and quadratic polynomials c2x2 + c1x + c0.

So, let's dive in and learn how to divide polynomials using synthetic division!

How to do synthetic division? Linear monic divisors

This section describes how to do synthetic division if the divisor is of the form x - b. For examples of synthetic division with divisors of a more complicated form, see the subsequent sections.

💡 It was Paolo Ruffini who described such division back in 1804. That's why you can sometimes encounter the term Ruffini's rule instead of synthetic division.

Since synthetic division is best explained with an example, we'll divide 3x3 - 8x - 9 by x - 2. Let's discuss in detail how to do synthetic division.

  1. Set up the synthetic division table. It consists of three rows:

    • In the first row, put the coefficients of the dividend in descending powers of x, inserting 0's for any missing powers. In our example, the x2 term is missing, so we add 0 between 3 and -8, i.e., between the coefficients of x3 and x.

    • In the second row and one column to the left, write b from the divisor x - b. In our case, b = 2. It is common to write the multiplication sign in front of b and to separate it from the coefficients with a vertical bar. This is because the role this number plays in synthetic division is different than that of the coefficients.

    • We leave the third row blank - we'll fill it up as we go.

⋅2[30−8−9]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ && & \\ && & \end{bmatrix}
  1. Drop the leading coefficient of the dividend to the bottom row.
⋅2[30−8−93]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ && & \\ \textcolor{green}{3} && & \end{bmatrix}
  1. Multiply this dropped number by the number b on the left (in our case, it's multiplication by 2). Place the result under the next coefficient of the dividend.
⋅2[30−8−963]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ &\textcolor{green}{6}& & \\ 3 && & \end{bmatrix}
  1. Sum the numbers in the column we created in the previous step. Write the result in the bottom row.
⋅2[30−8−9636]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ & 6& & \\ 3 &\textcolor{green}{6}& & \end{bmatrix}
  1. Repeat Steps 3. and 4. until the table is full. We'll show you how to do it:
  • Multiply the 6 we obtained above by b = 2 and write the result in the next column.
⋅2[30−8−961236]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ & 6& \textcolor{green}{12} & \\ 3 &6& & \end{bmatrix}
  • Add -8 and 12 together and write the sum in the last row.
⋅2[30−8−9612364]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ & 6& 12 & \\ 3 &6& \textcolor{green}{4}& \end{bmatrix}
  • Multiply the 4 we obtained above by b = 2 and write the result in the next column.
⋅2[30−8−96128364]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ & 6& 12 & \textcolor{green}{8}\\ 3 &6& 4& \end{bmatrix}
  • Add -9 and 8 together and place the result in the last row.
⋅2[30−8−96128364−1]\small \qquad \begin{array}{c} \\ \bold{\cdot 2} \\ \\ \end{array} \begin{bmatrix} 3& 0 &-8& -9\\ & 6& 12 &8\\ 3 &6& 4& \textcolor{green}{-1} \end{bmatrix}
  1. OK, the table is full. The result of the polynomial division we're looking for is here: the coefficients of the quotient and the remainder of this division can be found in the last row of our table!
364−1\small \qquad \begin{array}{c} \textcolor{red}{3} & \textcolor{red}{6} & \textcolor{red}{4} & \textcolor{blue}{-1} \end{array}

The last value on the right is the remainder, and all the other values are the consecutive coefficients of the quotient, starting from the leading coefficient (working left to right):

  • Coefficients of the quotient: 3, 6, 4
  • Quotient: 3x2 + 6x + 4
  • Remainder: -1

Synthetic division for non-monic linear divisors

We will now see how to perform a synthetic division if the divisor is in the form b1x + b0, i.e., linear but not necessarily monic. As an example, let's divide 4x3 + 2x2 - 2x + 1 by 2x + 1.

  1. Set up the division table. It's very similar to that for monic divisors, but the main difference is the presence of the fourth row, where we place the leading coefficient b1 of the divisor preceded by the division sign :. Also, be careful with the signs: we write -b0 at the beginning of the second row!
⋅(−1):2[42−21]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & & & \\ & & & \\ & & \end{bmatrix}
  1. Drop the leading coefficient of the dividend to the third row.
⋅(−1):2[42−214]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & & & \\ \textcolor{green}{4}& & & \\ & & \end{bmatrix}
  1. Divide the number you've just placed in the third row by 2 and put the result in the last row.
⋅(−1):2[42−2142]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & & & \\ 4& & & \\ \textcolor{green}{2} & & \end{bmatrix}
  1. Multiply this dropped number by the number on the left. Place the result under the next coefficient of the dividend.
⋅(−1):2[42−21−242]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & \textcolor{green}{-2}& & \\ 4& & & \\ 2 & & \end{bmatrix}
  1. Sum the numbers in the column we created in the previous step. Write the result in the third row.
⋅(−1):2[42−21−2402]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & -2 & & \\ 4& \textcolor{green}{0}& & \\ 2 & & \end{bmatrix}

  1. Repeat Steps 3. & 4. & 5. until the table is (almost) full – you don't need to perform the last division.

    In our example, we obtain the following table (check it yourself!):

⋅(−1):2[42−21−20140−2220−1]\small \quad \begin{array}{c} \\ \bold{\cdot (-1)} \\ \\ \bold{:2} \end{array} \begin{bmatrix} 4 &2 &-2& 1 \\ & -2 &0 &1\\ 4& 0& -2 &2 \\ 2 & 0 & -1 \end{bmatrix}
  1. This time, to find the coefficients of the quotient and the remainder, you need to take a look at the last two rows of our table.
40−2220−1\small \qquad \begin{array}{c} 4 & 0& -2& \textcolor{blue}{2} \\ \textcolor{red}{2} & \textcolor{red}{0} & \textcolor{red}{-1} & \end{array}

The last value of the penultimate row is the remainder of the division, and the values in the last row are the coefficients of the quotient: working left to right, the first number is the leading coefficient of the quotient, and the last one is the constant.

  • Coefficients of the quotient: 2, 0, -1
  • Quotient: 2x2 - 1
  • Remainder: 2

Synthetic division for higher-degree divisors

We will now see how to perform a synthetic division if the divisor is a quadratic polynomial of the form c2x2 + c1x + c0. As an example, let's divide 6x4 + 7x2 - 3 by 2x2 - 4x + 5.

  1. Set up the division table. Similarly to the non-monic linear case, the leading coefficient of the divisor, c2 (preceded by the division sign /), is placed in the last row. However, we need two rows for two non-leading coefficients of the divisor: we put -c0 in the second row, and -c1 in the third row and to the left so that these two numbers form a downward left diagonal. And, as always, be careful with the signs!
⋅(−5)⋅4:2[6070−3]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & & & \\ & & & & \\ & & & & \\ & & && \end{bmatrix}
  1. Drop the leading coefficient of the dividend to the penultimate row.
⋅(−5)⋅4:2[6070−36]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & & & \\ & & & & \\ \textcolor{green}{6}& & & & \\ & & && \end{bmatrix}
  1. Divide the number you've dropped by 2 and put the result into the last row.
⋅(−5)⋅4:2[6070−363]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & & & \\ & & & & \\ 6& & & & \\ \textcolor{green}{3}& & && \end{bmatrix}
  1. Multiply the result from the previous step by the numbers before the bar and place the results diagonally: the result of multiplying by 4 goes under the second coefficient, and the result of multiplying by -5 goes under the third coefficient.
⋅(−5)⋅4:2[6070−3−151263]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & \textcolor{green}{-15}& & \\ & \textcolor{green}{12}& & & \\ 6& & & & \\ 3& & && \end{bmatrix}
  1. Sum the numbers in the column we created in the second column and put the result into the penultimate row.
⋅(−5)⋅4:2[6070−3−15126123]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & -15& & \\ & 12& & & \\ 6& \textcolor{green}{12}& & & \\ 3& & && \end{bmatrix}
  1. Repeat Steps 3. & 4. & 5. until the penultimate row is full, i.e.:
  • Divide 12 from the previous step by 2 and put the result into the last row.
⋅(−5)⋅4:2[6070−3−151261236]\footnotesize\quad \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & -15& & \\ & 12& & & \\ 6& 12& & & \\ 3& \textcolor{green}{6}& && \end{bmatrix}
  • Multiply 6 by the numbers before the bar and place the results diagonally. The result of multiplying by 4 goes under the third coefficient, and the result of multiplying by -5 goes under the fourth coefficient.
 ⋅(−5)⋅4:2[6070−3−15−30122461236]\footnotesize\ \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & -15& \textcolor{green}{-30}& \\ & 12& \textcolor{green}{24}& & \\ 6& 12& & & \\ 3& 6& && \end{bmatrix}
  • Sum the numbers in the third column and put the result in the penultimate row.
 ⋅(−5)⋅4:2[6070−3−15−3012246121636]\footnotesize\ \begin{array}{cc} & \\ & \bold{\cdot (-5)} \\ \bold{\cdot 4} &\\ & \\ & \bold{:2} \end{array}\!\!\!\! \begin{bmatrix} 6& 0& 7& 0& -3 \\ & & -15& -30& \\ & 12& 24& & \\ 6& 12& \textcolor{green}{16} & & \\ 3& 6& && \end{bmatrix}
  • Divide 16 by 2 and put the result into the last row.