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

Completing the Square

Example 02: Complete the square for the polynomial $x^2+3x-1$$x^2+3x-1$.

Solution 02a:

$x^2+3x-1$$x^2+3x-1$ implies $a=1$$a=1$, $b=3$$b=3$, and $c=-1$$c=-1$.

$x^2+3x+0-1\Rightarrow x^2+3x+\mathrm{\_}\mathrm{\_}\mathrm{\_}-1-\mathrm{\_}\mathrm{\_}\mathrm{\_}$$x^2+3x\,{\color{red}+\,0}-1\Rightarrow x^2+3x\,{\color{red}+\,\_\_\_}-1\,{\color{red}-\,\_\_\_}$

To compute the value to replace the red underscores above:

${\displaystyle {\left({\displaystyle \frac{b}{2}}\right)}^2={\left({\displaystyle \frac{3}{2}}\right)}^2={\displaystyle \frac{9}{4}}}$$\displaystyle \left(\dfrac{b}{2}\right)^2=\left({\color{orange}\dfrac{3}{2}}\right)^2=\dfrac{9}{4}$

So,

$\Rightarrow x^2+3x+{\displaystyle \frac{9}{4}}-1-{\displaystyle \frac{9}{4}}$$\Rightarrow x^2+3x\,{\color{red}+\,\dfrac{9}{4}}-1\,{\color{red}-\,\dfrac{9}{4}}$

The first three terms above is a complete square:

$\Rightarrow {\left(x+{\displaystyle \frac{3}{2}}\right)}^2-1-{\displaystyle \frac{9}{4}}$$\Rightarrow \left(x{\color{orange}\,+\,\dfrac{3}{2}}\right)^2-1\,{\color{red}-\,\dfrac{9}{4}}$

Rewrite:

$\Rightarrow {\left(x+{\displaystyle \frac{3}{2}}\right)}^2-1-{\displaystyle \frac{9}{4}}\Rightarrow {(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{13}{4}}$$\Rightarrow \left(x{\color{orange}\,+\,\dfrac{3}{2}}\right)^2-1\,{\color{red}-\,\dfrac{9}{4}}\Rightarrow \left(x+\dfrac{3}{2}\right)^2-\dfrac{13}{4}$

Alternatively, use the formula $a{x}^2+bx+c=a{(x+{\displaystyle \frac{b}{2a}})}^2+(c-{\displaystyle \frac{b^2}{4a}})$$ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2+\left(c-\dfrac{b^2}{4a}\right)$

Solution 02b:

$x^2+3x-1$$x^2+3x-1$ implies $a=1$$a=1$, $b=3$$b=3$, and $c=-1$$c=-1$.

$1x^2+3x-1=1{(x+{\displaystyle \frac{3}{2(1)}})}^2+(-1-{\displaystyle \frac{(3{)}^2}{4(1)}})\Rightarrow {(x+{\displaystyle \frac{3}{2}})}^2-{\displaystyle \frac{13}{4}}$$1x^2+3x-1=1\left(x+\dfrac{3}{2(1)}\right)^2+\left(-1-\dfrac{(3)^2}{4(1)}\right)\Rightarrow \left(x+\dfrac{3}{2}\right)^2-\dfrac{13}{4}$

Visualization of Completing the Square for Even b

Check Your Work

Use Technology to Complete the Square

NOTE: On Assessments, you may use of technology to complete the square without showing any work.

*Mathematica

1
(* Complete the square for x^2+3x-1 *)
2
ResourceFunction["CompleteSquare"][x^2+3x-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. ResourceFunction[ ][ ] requires

   1. A call to the specific resource function using the first brackets, ["CompleteSquare"]

   2. The argument to apply the resource function to, [x^2+3x-1, x] For this resource function, the argument has two parts:

      1. The expression to complete the square, $\text{x}\wedge \text{2+3x-1}$$\text{x}\!\wedge\!\text{2+3x-1}$
      2. The independent variable, $\text{x}$$\text{x}$

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

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

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Monday, 06 September 2021 - 12:36 (EDT) Last Modified: Wednesday, 26 January 2022 - 04:06 (EST)