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

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

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

Isn't there a better way?

### Synthetic Division of Polynomials

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, \begin{align*}x - k\end{align*}. However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.

### Examples

#### Example 1

Earlier, you were given a question about finding a better way to divide polynomials.

If we want to use synthetic division, notice that the factor is not in the form

\begin{align*}(x - k)\end{align*}. Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If \begin{align*}x^2 + 2 = 0\end{align*}, then \begin{align*}x = \pm i \sqrt{2}\end{align*}, which can also be expressed as \begin{align*}x = +i \sqrt{2}\end{align*} or \begin{align*}x = -i \sqrt{2}\end{align*}. Therefore, we need to use synthetic division twice because there are two complex roots.

To start, put \begin{align*}i \sqrt{2}\end{align*} up in the left-hand corner box.

When we perform the synthetic division, we get a remainder of \begin{align*}-5 + 5 i \sqrt{2}\end{align*}.

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

As a result, \begin{align*}f(x) = \frac{3x^3 - 2x^2 + 11x - 9}{x^2 + 2} = 3x - 2 + \frac{5x - 5}{x^2 + 2}\end{align*}.

#### Example 2

Divide \begin{align*}2x^4-5x^3-14x^2+47x-30\end{align*} by \begin{align*}x - 2\end{align*}.

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

#### Example 3

Determine if 4 is a solution to \begin{align*}f(x)=5x^3+6x^2-24x-16\end{align*}.

Using synthetic division, we have:

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

**Remainder Theorem:** If \begin{align*}f(k) = r\end{align*}, then \begin{align*}r\end{align*} is also the remainder when dividing by \begin{align*}(x - k)\end{align*}.

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

#### Example 4

Determine if \begin{align*}(2x - 5)\end{align*} is a factor of \begin{align*}4x^4-9x^2-100\end{align*}.

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

This means that \begin{align*}\frac{5}{2}\end{align*} is a zero and its corresponding binomial, \begin{align*}(2x - 5)\end{align*}, is a factor.

#### Example 5

Is 6 a solution for \begin{align*}f(x)=x^3-8x^2+72\end{align*}? If so, find the real-number zeros (solutions) of the resulting polynomial.

Put a zero placeholder for the \begin{align*}x-\end{align*}term. Divide by 6.

The resulting polynomial is \begin{align*}x^2-2x-12\end{align*}. While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.

\begin{align*}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}\end{align*}

The solutions to this polynomial are 6, \begin{align*}1+\sqrt{13} \approx 4.61\end{align*} and \begin{align*}1-\sqrt{13} \approx -2.61\end{align*}.

#### Example 6

Divide \begin{align*}4x^3+3x^2+10x+4\end{align*} by \begin{align*}(x^2 + 2)\end{align*}. Write the resulting polynomial with the remainder (if there is one).

Using synthetic division, divide by \begin{align*}i \sqrt{2}\end{align*}.

Divide the results from the last step by \begin{align*}-i \sqrt{2}\end{align*}.

The answer is \begin{align*}4x + 3 + \frac{2x - 2}{x^2 + 2}\end{align*}.

### Review

Divide using synthetic division:

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

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

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

Divide using synthetic division:

- \begin{align*}4x^3 - 3x^2 + x + 1\end{align*} divided by \begin{align*}(x^2 - 1)\end{align*}
- \begin{align*}x^4 - x^2 + 1\end{align*} divided by \begin{align*}(x^2 + 1)\end{align*}
- \begin{align*}2x^4 - \frac{1}{2}x^2 + 2x\end{align*} divided by \begin{align*}(4x^2 - 1)\end{align*}
- \begin{align*}4x^3 - 3x^2 + x + 1\end{align*} divided by \begin{align*}(4x^2 + 1)\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.11.