Partial Fractions

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.4 – Integration of Rational Functions by Partial Fractions

Prerequisite Knowledge

This section contains the specific content from previous courses that you may need for CPT_06_Partial_Fractions.

Algebra

Example 01: Divide ${\displaystyle {\displaystyle \frac{x^3+2}{x+1}}}$$\displaystyle \dfrac{x^3+2}{x+1}$

Solution 01a:

Use Polynomial Division

Rewrite numerator will all missing terms:

${\displaystyle {\displaystyle \frac{x^3+2}{x+1}}={\displaystyle \frac{x^3+0x^2+0x+2}{x+1}}}$$\displaystyle \dfrac{x^3+2}{x+1}=\dfrac{x^3{\color{red}+0x^2+0x}+2}{x+1}$

Divide the dividend, $x^3+0x^2+0x+2$$x^3+0x^2+0x+2$, by the divisor, $x+1$$x+1$:

$\begin{array}{c}\phantom{xxxxxxxxxxxxx}{x}^2-x+1\\ x+1\surd \overline{x^3+0x^2+0x+2}\\ -(x^3+x^2)\\ \phantom{xxxxxx}\overline{\phantom{-xx{x}^3}-x^2-0x+2}\\ \phantom{xxxxxxx}-(-x^2-x)\\ \phantom{xxxxxx}\overline{\phantom{-xx{x}^3-x^2-}x+2}\\ \phantom{xxxxxxxxxxxxxxxxx}-(x+1)\\ \phantom{xxxxxx}\overline{\phantom{-xx{x}^3-x^2-x+}1}\end{array}$$\begin{array}{c} \phantom{xxxxxxxxxxxxx}{\color{red}x^2}\,\,{\color{green}-\,\,x}\,\,+\,\,{\color{purple}1}\\ x\,\,+\,\,1\,\,\surd\overline{x^3\,\,+\,\,0x^2\,\,+\,\,0x\,\,+\,\,2} \\ \!\!\!\!\!\!\!\!\!\!\!-({\color{red}x^3\,\,+\,\,x^2})\\ \phantom{xxxxxx}\overline{\phantom{-xxx^3}-x^2\,\,-\,\,0x\,\,+\,\,2}\\ \phantom{xxxxxxx}\!\!\!\!\!-({\color{green}-x^2\,\,-\,\,x})\\ \phantom{xxxxxx}\overline{\phantom{-xxx^3-x^2\,\,-\,\,}x\,\,+\,\,2} \\ \phantom{xxxxxxxxxxxxxxxxx}\!\!\!\!\!-({\color{purple}x\,\,+\,\,1})\\ \phantom{xxxxxx}\overline{\phantom{-xxx^3-x^2\,\,-\,\,x\,\,+\,\,}1} \end{array}$

The quotient is $x^2-x+1$$\color{orange}x^2-x+1$ and the remainder is $1$$\color{brown}1$.

Thus,

${\displaystyle {\displaystyle \frac{x^3+2}{x+1}}=x^2-x+1+{\displaystyle \frac{1}{x+1}}}$$\displaystyle \dfrac{x^3+2}{x+1}={\color{orange}x^2-x+1}+\dfrac{\color{brown}1}{x+1}$

Solution 01b:

Use Synthetic Division

Rewrite numerator will all missing terms:

${\displaystyle {\displaystyle \frac{x^3+2}{x+1}}={\displaystyle \frac{x^3+0x^2+0x+2}{x+1}}}$$\displaystyle \dfrac{x^3+2}{x+1}=\dfrac{x^3{\color{red}+0x^2+0x}+2}{x+1}$

Divide the dividend, $x^3+0x^2+0x+2$$x^3+0x^2+0x+2$, by the divisor, $x+1$$x+1$:

$\begin{array}{c}\phantom{xxxxxxxx}1\phantom{x}-1\phantom{xl}1\\ -1|\overline{1\phantom{xx}0\phantom{xll}0\phantom{xx}2}\\ \phantom{xxxx}\downarrow \phantom{l}-1\phantom{xll}1\phantom{xx}1\\ \phantom{xxxx}\overline{1\phantom{l}-1\phantom{xll}1\phantom{xx}1}\end{array}$$\begin{array}{c} \phantom{xxxxxxxx}{\color{red}1}\phantom{x}{\color{green}-1}\phantom{xl}{\color{purple}1}\\ -\,\,1\,\,\ \left|\overline{1\phantom{xx}0\phantom{xll}0\phantom{xx}2}\right. \\ \phantom{xxxx}\downarrow\phantom{l}{\color{red}-1}\phantom{xll}{\color{green}1}\phantom{xx}{\color{purple}1}\\ \phantom{xxxx}\overline{{\color{orange}\,1\phantom{l}-\!1\phantom{xll}1}\phantom{xx}\color{brown}1} \end{array}$

The quotient is $x^2-x+1$$\color{orange}x^2-x+1$ and the remainder is $1$$\color{brown}1$.

Thus,

${\displaystyle {\displaystyle \frac{x^3+2}{x+1}}=x^2-x+1+{\displaystyle \frac{1}{x+1}}}$$\displaystyle \dfrac{x^3+2}{x+1}={\color{orange}x^2-x+1}+\dfrac{\color{brown}1}{x+1}$

Use Technology to Perform Polynomial Division

NOTE: On Assessments, you may use of technology to perform polynomial division without showing any work.

Mathematica

1
(* Divide x^3+2 by x+1*)
2
PolynomialQuotientRemainder[x^3+2, x+1, x]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: https://www.jjw3.com/Common_Mathematica_Code.html.

2. PolynomialQuotientRemainder[ ] has three arguments:

   1. The function in the dividend (i.e., numerator), $\text{x}\wedge \text{3+2}$$\text{x}\!\wedge\!\text{3+2}$;
   2. The function in the divisor (i.e., denominator), $\text{x+1}$$\text{x+1}$;
   3. The independent variable.

3. For help on using the PolynomialQuotientRemainder[ ] function:

   1. In Mathematica, execute the code: $\text{?PolynomialQuotientRemainder}$$\text{?PolynomialQuotientRemainder}$
   2. Click on $\vee$$\or$ near the bottom-left of output
   3. Click on local
   4. Read how to use the PolynomialQuotientRemainder[ ] function – you will be able to copy-paste code.

4. Remember, correct Mathematica code will be all black except for variables.

5. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

6. Mathematica returns the result $\{1-x+x^2,1\}$$\{1-x+x^2,1\}$ in the form $\{\text{Quotient, Remainder}\}$$\{\text{Quotient, Remainder}\}$.

Python

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Python syntax is located at: https://www.jjw3.com/Common_Python_Code.html.
2. Python requires $\ast$$*$ for explicit multiplication.
3. Python requires $\ast \ast$$**$ for exponentiation, e.g., $x^3$$x^3$ is $\text{x**3}$$\text{x**3}$.
4. Remember to use parens $($$\left(\right.$ and $)$$\left.\right)$​ to group factors consisting of multiple terms.
5. You can use pprint(factor(8x*3-1)) for pretty print.
6. Python returns the result $(x\ast \ast 2-x+1,1)$$(x**2-x+1, 1)$ in the form $(\text{Quotient, Remainder})$$(\text{Quotient, Remainder})$.
7. To execute code, press the “View the result” button:

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Created: Monday, 06 September 2021 - 12:36 (EDT)