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Algebra 1

6.3.3 Dividing Polynomials Using Synthetic Division

Algebra 16.3.3 Dividing Polynomials Using Synthetic Division

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Activity

In this activity we’re going to use synthetic division to divide a polynomial by a binomial.

Part 1: Use Long Division

You are going to solve a division problem using long division.

1. ( x 2 13 x + 42 ) ÷ ( x 6 ) ( x 2 13 x + 42 ) ÷ ( x 6 )

Part 2: Use Synthetic Division

You are now going to solve the same division problem using synthetic division.

2. ( x 2 13 x + 42 ) ÷ ( x 6 ) ( x 2 13 x + 42 ) ÷ ( x 6 )

Work with a partner to find each quotient. Discuss any differences you have in determining the correct solution.

3. ( 4 x 3 + 5 x 2 5 x + 3 ) ÷ ( x + 2 ) ( 4 x 3 + 5 x 2 5 x + 3 ) ÷ ( x + 2 )

4. ( 2 x 3 11 x 2 + 16 x 12 ) ÷ ( x 4 ) ( 2 x 3 11 x 2 + 16 x 12 ) ÷ ( x 4 )

5. ( x 4 9 x 2 + 2 x + 6 ) ÷ ( x + 3 ) ( x 4 9 x 2 + 2 x + 6 ) ÷ ( x + 3 )

6. ( 3 x 4 11 x 3 + 2 x 2 + 10 x + 6 ) ÷ ( x 3 ) ( 3 x 4 11 x 3 + 2 x 2 + 10 x + 6 ) ÷ ( x 3 )

Self Check

Use synthetic division to find the quotient and remainder when x 4 + x 2 + 6 x 10 is divided by x + 2 .
  1. x 3 2 x 2 + 5 x 4 ; remainder of 2
  2. x 3 2 x 2 + 5 x 4 ; remainder of 2
  3. x 3 2 x 2 + 5 x 4 ; remainder of 0
  4. x 3 x 2 + 8 x ; remainder of 26

Additional Resources

Dividing Polynomials Using Synthetic Division

Mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in a previous example and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.

A polynomial long division and synthetic division of x² + 9x + 20 by x + 5, showing matching coefficients and the remainder, with colored numbers highlighting key steps and results.

Synthetic division basically just removes unnecessary repeated variables and numbers. Here, all the x x and x 2 x 2 are removed as well as the x 2 x 2 and 4 x 4 x because they are opposite the term above.

The first row of the synthetic division is the coefficients of the dividend. The 5 5 is the opposite of the 5 5 in the divisor.

The second row of the synthetic division is the numbers shown in red in the division problem.

The third row of the synthetic division is the numbers shown in blue in the division problem.

Notice the quotient and remainder are shown in the third row.

Synthetic division only works when the divisor is of the form x c x c .

The following example will explain the process.

Example 1

Use synthetic division to find the quotient and remainder when 2 x 3 + 3 x 2 + x + 8 2 x 3 + 3 x 2 + x + 8 is divided by x + 2 x + 2 .

Step 1 - Write the dividend with decreasing powers of x x .

2 x 3 + 3 x 2 + x + 8 2 x 3 + 3 x 2 + x + 8

Step 2 - Write the coefficients of the terms as the first row of the synthetic division.

Synthetic division setup: blank  outside, coefficients 2, 3, 1, and 8 inside.

Step 3 - Write the divisor as x c x c and place c c in the synthetic division in the divisor box.

Synthetic division setup -2 outside, coefficients 2, 3, 1, and 8 inside.

Step 4 - Bring down the first coefficient to the third row.

Synthetic division setup -2 outside, coefficients 2, 3, 1, and 8 inside. in the top row, and 2 written below the line under the 2, with a downward green arrow pointing from the top 2 to the bottom 2.

Step 5 - Multiply that coefficient by the divisor and place the result in the second row under the second coefficient.

A synthetic division setup: -2 on the left, with 2, 3, 1, and 8 in a row to the right. Two arrows show bringing down 2 and multiplying it by -2 to get -4, placed below the 3.

Step 6 - Add the second column, putting the result in the third row.

Synthetic division setup showing the divisor -2 and dividend 2, 3, 1, 8. The first calculation, 2 multiplied by -2 equals -4, written under the 3 is shown with an arrow pointing down to -1., representing 3 minus 4.

Step 7 - Multiply that result by the divisor and place the result in the second row under the third coefficient.

Synthetic division setup: -2 outside, coefficients 2, 3, 1, and 8 inside. Green arrows indicate -2 times -1 equal to 2 that is placed under the 1..

Step 8 - Add the third column, putting the result in the third row.

Synthetic division setup: -2 outside, coefficients 2, 3, 1, and 8 inside. A green arrow indicates 2 plus 1 equal to 3 that is placed under the 2.

Step 9 - Multiply that result by the divisor and place the result in the third row under the third coefficient.

Synthetic division setup showing -2 as the divisor and 2, 3, 1, 8 as the coefficients. Arrows indicate the process of multiplying and adding during the division steps. Green arrows go from -2 to 3 and 3 to -6 placed under the 8.  The negative 6 represents -2 times 3.

Step 10 - Add the final column, putting the result in the third row.

Synthetic division setup with -2 as the divisor and 2, 3, 1, 8 as the coefficients. The steps show the intermediate results: -4, 2, -6 and the final bottom row: 2, -1, 3, 2. A green arrow goes from 8 to -6 to the 2.  It represents 8-6=2.

Step 11 - The quotient is 2 x 2 1 x + 3 2 x 2 1 x + 3 , and the remainder is 2 2 .

The division is complete. The numbers in the third row give us the result. The 2 2 1 1 3 3 are the coefficients of the quotient. The quotient is 2 x 2 1 x + 3 2 x 2 1 x + 3 . The 2 2 in the box in the third row is the remainder.

Compare your answer:

(quotient)(divisor) + remainder = dividend ( 2 x 2 1 x + 3 ) ( x + 2 ) + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 x 2 + 3 x + 4 x 2 2 x + 6 + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 + 3 x 2 + x + 8 = 2 x 3 + 3 x 2 + x + 8 ✓ (quotient)(divisor) + remainder = dividend ( 2 x 2 1 x + 3 ) ( x + 2 ) + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 x 2 + 3 x + 4 x 2 2 x + 6 + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 + 3 x 2 + x + 8 = 2 x 3 + 3 x 2 + x + 8 ✓ (quotient)(divisor) + remainder = dividend ( 2 x 2 1 x + 3 ) ( x + 2 ) + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 x 2 + 3 x + 4 x 2 2 x + 6 + 2 = ? 2 x 3 + 3 x 2 + x + 8 2 x 3 + 3 x 2 + x + 8 = 2 x 3 + 3 x 2 + x + 8 ✓

Example 2

Use synthetic division to find the quotient and remainder when x 4 16 x 2 + 3 x + 12 x 4 16 x 2 + 3 x + 12 is divided by x + 4 x + 4 .

The polynomial x 4 16 x 2 + 3 x + 12 x 4 16 x 2 + 3 x + 12 has its terms in order with descending degree, but we notice there is no x 3 x 3 term. We will add a 0 0 as a placeholder for the x 3 x 3 term. In x c x c form, the divisor is x ( 4 ) x ( 4 ) .

A synthetic division setup shows the number -4 on the left and the top row with coefficients: 1, 0, -16, 3, 12. The bottom two rows have numbers: -4, 16, 0, -12; and 1, -4, 0, 3, 0.

We divided a 4th degree polynomial by a 1st degree polynomial, so the quotient will be a 3rd degree polynomial. Reading from the third row, the quotient has the coefficients 1 1 4 4 0 0 3 3 , which is x 3 4 x 2 + 3 x 3 4 x 2 + 3 . The remainder is 0 0 .

Try it

Try It: Dividing Polynomials Using Synthetic Division

1. Use synthetic division to find the quotient and remainder when 3 x 3 + 10 x 2 + 6 x 2 3 x 3 + 10 x 2 + 6 x 2 is divided by x + 2 x + 2 .

2. Use synthetic division to find the quotient and remainder when x 4 16 x 2 + 5 x + 20 x 4 16 x 2 + 5 x + 20 is divided by x + 4 x + 4 .

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