# 6.10: Synthetic Division of Polynomials

**At Grade**Created by: CK-12

**Practice**Synthetic Division of Polynomials

The volume of a rectangular prism is . Determine if is the length of one of the prism's sides.

### Watch This

James Sousa: Polynomial Division: Synthetic Division

### Guidance

Synthetic division is an alternative to long division from the previous concept. It can also be used to divide a polynomial by a possible factor, . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

#### Example A

Divide by .

**
Solution:
**
Using synthetic division, the setup is as follows:

To “read” the answer, use the numbers as follows:

Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is . Notice that when we synthetically divide by , the “leftover” polynomial is one degree less than the original. We could also write .

#### Example B

Determine if 4 is a solution to .

Using synthetic division, we have:

The remainder is 304, so 4 is not a solution. Notice if we substitute in , also written , we would have . This leads us to the Remainder Theorem.

**
Remainder Theorem:
**
If
, then
is also the remainder when dividing by
.

This means that if you substitute in or divide by , what comes out of is the same. is the remainder, but also is the corresponding value. Therefore, the point would be on the graph of .

#### Example C

Determine if is a factor of .

**
Solution:
**
If you use synthetic division, the factor is not in the form
. We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put
up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this example, we need zeros for the
term and the
term.

This means that is a zero and its corresponding binomial, , is a factor.

**
Intro Problem Revisit
**

If divides evenly into then it is the length of one of the prism's sides.

If we use synthetic division, the factor is not in the form . We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of 0. This means that is a factor of the volume. Therefore, it is also the length of one of the sides of the rectangular prism.

### Guided Practice

1. Divide by . Write the resulting polynomial with the remainder (if there is one).

2. Divide by . Write the resulting polynomial with the remainder (if there is one).

3. Is 6 a solution for ? If so, find the real-number zeros (solutions) of the resulting polynomial.

#### Answers

1. Using synthetic division, divide by -3.

The answer is .

2. Using synthetic division, divide by .

The answer is .

3. Put a zero placeholder for the term. Divide by 6.

The resulting polynomial is . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

The solutions to this polynomial are 6, and .

### Vocabulary

- Synthetic Division
- An alternative to long division for dividing by where only the coefficients of are used.

- Remainder Theorem
- If , then is also the remainder when dividing by .

### Practice

Use synthetic division to divide the following polynomials. Write out the remaining polynomial.

- Which of the division problems above generate no remainder? What does that mean?
- What is the difference between a zero and a factor?
- Find if .
- Now, divide by synthetically. What do you notice?

Find all real zeros of the following polynomials, given one zero.

Find all real zeros of the following polynomials, given two zeros.

### Image Attributions

## Description

## Learning Objectives

Here you'll learn how to use synthetic division as a short-cut and alternative to long division (in certain cases) and to find zeros.