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

*
Find the oblique asymptote of
f(x)=\frac{3x^{3}-2x^{2}+11x-9}{x^{2}+2}
*

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, @$x - k@$ . However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

#### Example A

Divide @$2x^4-5x^3-14x^2+47x-30@$ by @$x - 2@$ .

**
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 @$2x^3-x^2-16x+15@$ . Notice that when we synthetically divide by @$k@$ , the “leftover” polynomial is one degree less than the original. We could also write @$(x-2)(2x^3-x^2-16x+15)=2x^4-5x^3-14x^2+47x-30@$ .

#### Example B

Determine if 4 is a solution to @$f(x)=5x^3+6x^2-24x-16@$ .

Using synthetic division, we have:

The remainder is 304, so 4 is not a solution. Notice if we substitute in @$x = 4@$ , also written @$f(4)@$ , we would have @$f(4)=5(4)^3+6(4)^2-24(4)-16=304@$ . This leads us to the Remainder Theorem.

**
Remainder Theorem:
**
If
@$f(k) = r@$
, then
@$r@$
is also the remainder when dividing by
@$(x - k)@$
.

This means that if you substitute in @$x = k@$ or divide by @$k@$ , what comes out of @$f(x)@$ is the same. @$r@$ is the remainder, but it is also the corresponding @$y-@$ value. Therefore, the point @$(k, r)@$ would be on the graph of @$f(x)@$ .

#### Example C

Determine if @$(2x - 5)@$ is a factor of @$4x^4-9x^2-100@$ .

**
Solution:
**
If you use synthetic division, the factor is not in the form
@$(x - k)@$
. We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put
@$\frac{5}{2}@$
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
@$x^3-@$
term and the
@$x-@$
term.

This means that @$\frac{5}{2}@$ is a zero and its corresponding binomial, @$(2x - 5)@$ , is a factor.

#### Intro Problem Revisit

If we want to use synthetic division, notice that the factor is not in the form @$(x - k)@$ . Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If @$x^2 + 2 = 0@$ , then @$x = \pm i \sqrt{2}@$ , which can also be expressed as @$x = +i \sqrt{2}@$ or @$x = -i \sqrt{2}@$ . Therefore, we need to use synthetic division twice because there are two complex roots .

To start, put @$i \sqrt{2}@$ up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of @$-5 + 5 i \sqrt{2}@$ .

Next, we divide results from the last synthetic division with the other complex root. Put @$-i \sqrt{2}@$ up in the left-hand corner box.

As a result, @$f(x) = \frac{3x^3 - 2x^2 + 11x - 9}{x^2 + 2} = 3x - 2 + \frac{5x - 5}{x^2 + 2}@$ .

### Guided Practice

1. Divide @$x^3+9x^2+12x-27@$ by @$(x + 3)@$ . Write the resulting polynomial with the remainder (if there is one).

2. Is 6 a solution for @$f(x)=x^3-8x^2+72@$ ? If so, find the real-number zeros (solutions) of the resulting polynomial.

3. Divide @$4x^3+3x^2+10x+4@$ by @$(x^2 + 2)@$ . Write the resulting polynomial with the remainder (if there is one).

#### Answers

1. Using synthetic division, divide by @$-\frac{1}{2}@$ .

The answer is @$2x^3-12x^2+18x-\frac{2}{2x+1}@$ .

2. Put a zero placeholder for the @$x-@$ term. Divide by 6.

The resulting polynomial is @$x^2-2x-12@$ . While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

@$$x=\frac{2 \pm \sqrt{2^2-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}@$$

The solutions to this polynomial are 6, @$1+\sqrt{13} \approx 4.61@$ and @$1-\sqrt{13} \approx -2.61@$ .

3. Using synthetic division, divide by @$i \sqrt{2}@$ .

Divide the results from the last step by @$-i \sqrt{2}@$ .

The answer is @$4x + 3 + \frac{2x - 2}{x^2 + 2}@$ . -->

### Explore More

**
Divide using synthetic division:
**

- @$\frac{7x^2 - 23x + 6}{x - 3}@$
- @$\frac{x^4 - 5x + 10}{x + 3}@$
- @$(2x^2 + 13x - 8) \div (x - \frac{1}{2})@$
- @$(x^4 + 6x^3 + 6x^2) \div (x + 5)@$
- @$\frac{x^3 - 7x - 6}{x + 2}@$
- @$\frac{8y^3 + y^4 + 16 + 32y + 24y^2}{y + 2}@$

**
Use synthetic substitution to evaluate the polynomial function for the given value:
**

- @$P(x) = 2x^2 - 5x - 3@$ for @$x = 4@$
- @$P(x) = 4x^3 - 5x^2 + 3 @$ for @$x = -1@$
- @$p(x) = 3x^3 - 5x^2 - x =2@$ for @$x = -\frac{1}{3}@$
- The area of a rectangle is @$ 3x^3 - 11x^2 - 56x - 48@$ and the length is @$3x + 4@$ . 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 @$V(h) = \frac {1}{4} \pi h^3@$ . The mass (in grams) of each sample in terms of height can be modeled by @$M(h) = \frac{1}{4}h^3 - h^2 + 5h@$ . Write an expression that represents the density of the samples. (Hint: @$D = \frac{M}{V}@$ )

**
Divide using synthetic division:
**

- @$4x^3 - 3x^2 + x + 1@$ divided by @$(x^2 - 1)@$
- @$x^4 - x^2 + 1@$ divided by @$(x^2 + 1)@$
- @$2x^4 - \frac{1}{2}x^2 + 2x@$ divided by @$(4x^2 - 1)@$
- @$4x^3 - 3x^2 + x + 1@$ divided by @$(4x^2 + 1)@$