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?
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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,
Example A
Divide
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
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
Remainder Theorem: If
This means that if you substitute in
Example C
Determine if
Solution: If you use synthetic division, the factor is not in the form
This means that
Intro Problem Revisit
If we want to use synthetic division, notice that the factor is not in the form
To start, put
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
As a result,
Guided Practice
1. Divide
2. Is 6 a solution for
3. Divide
Answers
1. Using synthetic division, divide by
The answer is
2. Put a zero placeholder for the
The resulting polynomial is
The solutions to this polynomial are 6,
3. 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*}
Explore More
Divide using synthetic division:

\begin{align*}\frac{7x^2  23x + 6}{x  3}\end{align*}
7x2−23x+6x−3 
\begin{align*}\frac{x^4  5x + 10}{x + 3}\end{align*}
x4−5x+10x+3 
\begin{align*}(2x^2 + 13x  8) \div (x  \frac{1}{2})\end{align*}
(2x2+13x−8)÷(x−12) 
\begin{align*}(x^4 + 6x^3 + 6x^2) \div (x + 5)\end{align*}
(x4+6x3+6x2)÷(x+5) 
\begin{align*}\frac{x^3  7x  6}{x + 2}\end{align*}
x3−7x−6x+2 
\begin{align*}\frac{8y^3 + y^4 + 16 + 32y + 24y^2}{y + 2}\end{align*}
8y3+y4+16+32y+24y2y+2
Use synthetic substitution to evaluate the polynomial function for the given value:

\begin{align*}P(x) = 2x^2  5x  3\end{align*}
P(x)=2x2−5x−3 for \begin{align*}x = 4\end{align*}x=4 
\begin{align*}P(x) = 4x^3  5x^2 + 3 \end{align*}
P(x)=4x3−5x2+3 for \begin{align*}x = 1\end{align*}x=−1 
\begin{align*}p(x) = 3x^3  5x^2  x =2\end{align*}
p(x)=3x3−5x2−x=2 for \begin{align*}x = \frac{1}{3}\end{align*}x=−13  The area of a rectangle is \begin{align*} 3x^3  11x^2  56x  48\end{align*}
3x3−11x2−56x−48 and the length is \begin{align*}3x + 4\end{align*}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 \begin{align*}V(h) = \frac {1}{4} \pi h^3\end{align*}
V(h)=14πh3 . 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*}M(h)=14h3−h2+5h . Write an expression that represents the density of the samples. (Hint: \begin{align*}D = \frac{M}{V}\end{align*}D=MV )
Divide using synthetic division:

\begin{align*}4x^3  3x^2 + x + 1\end{align*}
4x3−3x2+x+1 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*}