If you have completed the lesson on oblique asymptotes, you probably know that the problem:

*
Find the oblique asymptote of
*

could be an excellent reason to groan and grumble due to the long division that would be required.

Isn't there a better way?

### 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 it is also the corresponding value. Therefore, the point would be on the graph of .

#### Example C

Determine if is a factor of .

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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 we want to use synthetic division, notice that the factor is not in the form . Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If , then , which can also be expressed as or . Therefore, we need to use synthetic division twice because there are two complex roots .

To start, put up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of .

Next, we divide results from the last synthetic division with the other complex root. Put up in the left-hand corner box.

As a result, .

### Guided Practice

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

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

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

#### Answers

1. Using synthetic division, divide by .

The answer is .

2. 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 .

3. Using synthetic division, divide by .

Divide the results from the last step by .

The answer is . -->

### Explore More

**
Divide using synthetic division:
**

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Use synthetic substitution to evaluate the polynomial function for the given value:
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- for
- for
- for
- The area of a rectangle is and the length is . What is the width?
- A group of geologists have taken a collection of samples of a substance from a proposed mining site and must identify the substance. Each sample is roughly cylindrical. The volume of each sample as a function of cylinder height (in centimeters) is . The mass (in grams) of each sample in terms of height can be modeled by . Write an expression that represents the density of the samples. (Hint: )

**
Divide using synthetic division:
**

- divided by
- divided by
- divided by
- divided by