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Precalculus for Team-Based Inquiry Learning 2025 Edition — Instructor Version

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Section 4.4 Polynomial Long Division (PR4)

Subsection 4.4.1 Activities

Observation 4.4.1.

We have seen previously that we can reduce rational functions by factoring, for example
\begin{equation*} \dfrac{x^2+5x+4}{x^3+3x^2+2x}=\dfrac{(x+1)(x+4)}{x(x+2)(x+1)}=\dfrac{x+4}{x(x+2)}. \end{equation*}
In this section, we will explore the question: what can we do to simplify rational functions if we are not able to reduce by easily factoring?
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Definition 4.4.2.

Recall that a fraction is called proper if its numerator is smaller than its denominator, and improper if the numerator is larger than the denominator (so \(\dfrac{3}{5}\) is a proper fraction, but \(\dfrac{32}{7}\) is an improper fraction). Similarly, we define a proper rational function to be a rational function where the degree of the numerator is less than the degree of the denominator.
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Activity 4.4.3.

Label each of the following rational functions as either proper or improper.
  1. \(\displaystyle \dfrac{x^3+x}{x^2+4}\)
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  2. \(\displaystyle \dfrac{3}{x^2+3x+4}\)
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  3. \(\displaystyle \dfrac{7+x^3}{x^2+x+1}\)
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  4. \(\displaystyle \dfrac{x^4+x+1}{x^4+4x^2}\)
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Answer.
A, C, and D are improper, while B is proper.
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Observation 4.4.4.

When dealing with an improper fraction such as \(\dfrac{32}{7}\text{,}\) it is sometimes useful to rewrite this as an integer plus a proper fraction, e.g. \(\dfrac{32}{7}=4+\dfrac{4}{7}\text{.}\) Similarly, it will sometimes be useful to rewrite an improper rational function as the sum of a polynomial and a proper rational function, such as \(\dfrac{x^3+x}{x^2+4}=x-\dfrac{3x}{x^2+4}\text{.}\)
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Activity 4.4.5.

\(\dfrac{357}{11}\text{.}\)
(a)
Use long division to write \(\dfrac{357}{11}\) as an integer plus a proper fraction.
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Answer.
So \(\dfrac{357}{11}=32+\dfrac{5}{11}\text{.}\)
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(b)
Now we will carefully redo this process in a way that we can generalize to rational functions. Note that we can rewrite \(357\) as \(357=3\cdot10^2+5\cdot10+7\text{,}\) and \(11\) as \(11=1\cdot10+1\text{.}\) By comparing the leading terms in these expansions, we see that to knock off the leading term of \(357\text{,}\) we need to multiply \(11\) by \(3\cdot10^1\text{.}\)
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Using the fact that \(357=11\cdot30+27\text{,}\) rewrite \(\dfrac{357}{11}\) as \(\dfrac{357}{11}=30+\dfrac{?}{11}\text{.}\)
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Answer.
\(\dfrac{357}{11}=30+\dfrac{27}{11}\text{.}\)
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(c)
Note now that if we can rewrite \(\dfrac{27}{11}\) as an integer plus a proper fraction, we will be done, since \(\dfrac{357}{11}=30+\dfrac{27}{11}\text{.}\)
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Rewrite \(\dfrac{27}{11}=?+\dfrac{?}{11}\) as an integer plus a proper fraction.
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Answer.
\(\dfrac{27}{11}=2+\dfrac{5}{11}\text{.}\)
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(d)
Combine your work in the previous two parts to rewrite \(\dfrac{357}{11}\) as an integer plus a proper fraction. How does this compare to what you obtained in part (a)?
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Answer.
\(\dfrac{357}{11}=30+\dfrac{27}{11}=30+2+\dfrac{5}{11}=32+\dfrac{5}{11}\text{.}\)
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Activity 4.4.6.

Now let’s consider the rational function \(\dfrac{3x^2+5x+7}{x+1}\text{.}\) We want to rewrite this as a polynomial plus a proper rational function.
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(a)
Looking at the leading terms, what do we need to multiply \(x+1\) by so that it would have the same leading term as \(3x^2+5x+7\text{?}\)
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  1. \(\displaystyle 3\)
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  2. \(\displaystyle x\)
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  3. \(\displaystyle 3x\)
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  4. \(\displaystyle 3x+5\)
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Answer.
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(b)
Rewrite \(3x^2+5x+7=3x(x+1)+?\text{,}\) and use this to rewrite \(\dfrac{3x^2+5x+7}{x+1}=3x+\dfrac{?}{x+1}\text{.}\)
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Answer.
\(\dfrac{3x^2+5x+7}{x+1}=3x+\dfrac{2x+7}{x+1}\)
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(c)
Now focusing on \(\dfrac{2x+7}{x+1}\text{,}\) what do we need to multiply \(x+1\) by so that it would have the same leading term as \(2x+7\text{?}\)
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  1. \(\displaystyle 2\)
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  2. \(\displaystyle x\)
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  3. \(\displaystyle 2x\)
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  4. \(\displaystyle 2x+7\)
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Answer.
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(d)
Rewrite \(\dfrac{2x+7}{x+1}=2+\dfrac{?}{x+1}\text{.}\)
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Answer.
\(\dfrac{2x+7}{x+1}=2+\dfrac{5}{x+1}\)
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(e)
Combine this with the previous parts to rewrite \(\dfrac{3x^2+5x+7}{x+1}=3x+?+\dfrac{?}{x+1}\text{.}\)
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Answer.
\(\dfrac{3x^2+5x+7}{x+1}=3x+2+\dfrac{5}{x+1}\)
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Activity 4.4.7.

Next we will use the notation of long division to rewrite the rational function \(\dfrac{3x^2+5x+7}{x+1}\) as a polynomial plus a proper rational function.
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(a)
First, let’s use long division notation to write the quotient.
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What do we need to multiply \(x+1\) by so that it would have the same leading term as \(3x^2+5x+7\text{?}\)
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Answer.
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(b)
Now to rewrite \(3x^2+5x+7\) as \(3x(x+1)+?\text{,}\) place the product \(3x(x+1)\) below and subtract.
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Answer.
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(c)
Now focusing on \(2x+7\text{,}\) what do we need to multiply \(x+1\) by so that it would have the same leading term as \(2x+7\text{?}\)
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Answer.
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(d)
Now, subtract \(2(x+1)\) to finish the long division.
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Answer.
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(e)
This long division calculation has shown that
\begin{equation*} 3x^2+5x+7 = (x+1)(3x+2)+5\text{.} \end{equation*}
Use this to rewrite \(\dfrac{3x^2+5x+7}{x+1}\) as a polynomial plus a proper rational function.
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Answer.
\(\dfrac{3x^2+5x+7}{x+1} = 3x+2+\dfrac{5}{x+1}\)
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Activity 4.4.9.

Rewrite \(\dfrac{x^2+1}{x-1}\) as a polynomial plus a proper rational function.
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Hint.
Note that \(x^2+1=x^2+0x+1\text{.}\)
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Answer.
\(x+1+\dfrac{3}{x-1}\text{.}\)
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Activity 4.4.10.

Rewrite \(\dfrac{x^5+x^3+2x^2-6x+7}{x^2+x-1}\) as a polynomial plus a proper rational function.
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Answer.
\(x^3-x^2+3x-2+\dfrac{-x+5}{x^2+x-1}\text{.}\)
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Activity 4.4.11.

Rewrite \(\dfrac{3x^4-5x^2+2}{x-1}\) as a polynomial plus a proper rational function.
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Answer.
\(3x^3+3x^2-2x-2\text{.}\)
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Subsection 4.4.2 Exercises

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