Synthetic division

Synthetic division is a method used to perform polynomial division. It can be used as a shortcut in place of algebraic long division when dividing a polynomial by a linear factor. It is important to note that it only works when dividing by a linear factor, such as (x + 1).

Synthetic division is typically used to test whether a value is a zero, or root, of a polynomial. If we divide a polynomial, such as x2 - 2x - 3, by one of its factors, we will get a remainder of 0. For example, we can factor x2 - 2x - 3 as:

x2 - 2x - 3 = (x + 1)(x - 3)

Based on this, we know that x = -1 and x = 3 are roots of the polynomial. In order to confirm this, we can divide x2 - 2x - 3 by (x + 1) and by (x - 3), and we should get a remainder of 0 in both cases. Use the following steps to perform synthetic division.

  1. Write only the coefficients of the polynomial inside an L-shaped symbol.
  2. Write just the root we are testing. If our factor is (x - 3), the root is x = +3. If our factor is (x + 1), the root is x = -1.
  3. From left to right, bring the first coefficient down, then multiply it by the root, and place the result underneath the following coefficient, then sum them.
  4. Multiply the result by the root, and write the result under the following coefficient, continuing this process through the last coefficient.
  5. The resulting numbers in the bottom row are the coefficients of the quotient. Count one from the right and separate the last term from the rest using a vertical bar. The last term is the remainder.
  6. Write the coefficients alongside the variables of the appropriate degree. The degree of the variables increases by 1 per term from right to left starting with a degree of 0 for the remainder and the last term before the remainder. If the remainder is not 0, write the value over the factor. For example, if the remainder is 12 and the factor is (x - 3), write .


Use synthetic division to divide x2 - 2x - 3 by (x + 1):







6. 1x - 3 = (x - 3)

Notice that this is the other factor, telling us that both (x - 3) and (x + 1) are factors of x2 - 2x - 3. If we were to divide the polynomial instead by (x - 3), we would get a quotent of (x + 1), also with a remainder of 0.