What is Synthetic Division?
Synthetic division is used to divide polynomials by first degree binomials.
A first degree binomial is an expression that has a variable that is not raised to any power. So, x + 5 is an example of a first degree binomial.
Free exam sample questions are provided at the end of this page.
Synthetic Division – Explanation of Terminology
To perform synthetic division, you need to identify the coefficients and constants in the polynomial.
You will remember that a polynomial is an algebraic expression that has two or more terms.
Terms can contain either numbers or variables or both.
A coefficient is the number part of a term.
A constant is a number on its own. Constants are usually placed at the end of an algebraic expression.
Synthetic Division – Example
Consider the following polynomial:
x4 + 5x3 – 2x2 – 28x – 12
In the second term, which is 5x3, the coefficient is 5, for example.
Now look at this practice problem.
Example: Divide the polynomial x4 + 5x3 – 2x2 – 28x – 12 by the first degree binomial x + 3. Use synthetic division to find your answer.
STEP 1: Identify the coefficients and constants.
Remember that if you have a variable without a number in front of it, the coefficient is 1.
x4 + 5x3 – 2x2 – 28x – 12 =
(1 × x4) + (5 × x3) – (2 × x2) – (28 × x) – 12 =
(1 × x4) + (5 × x3) + (–2 × x2) + (–28 × x) + (–12)
Our coefficients and constants are the numbers before each of the multiplication signs.
The coefficients are 1, 5, –2, and –28.
The constant is –12.
STEP 2: Identify the constant in the binomial and find its opposite.
We are going to divide by the binomial x + 3.
So, the constant in our binomial is 3.
The opposite of the constant is –3.
STEP 3: Write the problem out in synthetic division format, placing the opposite of the constant in an upside-down division sign.
Divide the numbers from step 1 by the opposite of the constant from step 2.
–3| 1 5 –2 –28 –12
STEP 4: Perform the synthetic division.
Bring down the 1 and multiply it by –3. Put that result under the 5. Then add 5 and –3 to get 2.
–3| 1 5 –2 –28 –12
–3 –6 24 12
1 2
STEP 5: Repeat the operations from step 4 until you find the remainder.
–3| 1 5 –2 –28 –12
–3 –6 24 12
1 2 –8 –4 0
Now try the practice problems in the next section.
Synthetic Division – Exercises
Instructions: Perform synthetic division to find the remainder for the following problems. If you do not know how to perform synthetic division, please see the example above before completing the exercises.
1) (x3 + 2x2 + x + 5) ÷ (x + 2)
2) (x3 – 6x2 – 2x + 14) ÷ (x – 6)
3) (x3 + x2 – x – 4) ÷ (x – 2)
4) (x4 + 5x3 + 10x2 + 14x + 10) ÷ (x + 3)
5) (x4 – 9x3 + 13x2 + 4x + 21) ÷ (x – 7)
Synthetic Division – Answers
1) The remainder is 3.
Our question was: (x3 + 2x2 + x + 5) ÷ (x + 2)
The opposite of the constant in our binomial is –2.
Our coefficients and constant are: 1 2 1 5
Set up the synthetic division to solve as shown below.
–2| 1 2 1 5
–2 0 –2
1 0 1 3
2) The remainder is 2.
Our question was: (x3 – 6x2 – 2x + 14) ÷ (x – 6)
The opposite of the constant in our binomial is 6.
Our coefficients and constant are: 1 –6 –2 14
Set up the synthetic division to solve as shown below.
6| 1 –6 –2 14
6 0 –12
1 0 –2 2
3) The remainder is 6.
Our question was: (x3 + x2 – x – 4) ÷ (x – 2)
The opposite of the constant in our binomial is 2.
Our coefficients and constant are: 1 1 –1 –4
Set up the synthetic division to solve as shown below.
2| 1 1 –1 –4
2 6 10
1 3 5 6
4) The remainder is 4.
Our question was: (x4 + 5x3 + 10x2 + 14x + 10) ÷ (x + 3)
The opposite of the constant in our binomial is –3.
Our coefficients and constant are: 1 5 10 14 10
Set up the synthetic division to solve as shown below.
–3| 1 5 10 14 10
–3 –6 –12 –6
1 2 4 2 4
5) The remainder is 0.
Our question was: (x4 – 9x3 + 13x2 + 4x + 21) ÷ (x – 7)
The opposite of the constant in our binomial is 7.
Our coefficients and constant are: 1 –9 13 4 21
Set up the synthetic division to solve as shown below.
7| 1 –9 13 4 21
7 –14 –7 –21
1 –2 –1 –3 0
You should also view our posts on the remainder theorem and factoring.