# 9.4 Partial fractions  (Page 2/7)

 Page 2 / 7

Given a rational expression with distinct linear factors in the denominator, decompose it.

1. Use a variable for the original numerators, usually $\text{\hspace{0.17em}}A,B\text{,\hspace{0.17em}}$ or $\text{\hspace{0.17em}}C,\text{\hspace{0.17em}}$ depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use $\text{\hspace{0.17em}}{A}_{n}\text{\hspace{0.17em}}$ for each numerator
$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

## Decomposing a rational function with distinct linear factors

Decompose the given rational expression with distinct linear factors.

$\frac{3x}{\left(x+2\right)\left(x-1\right)}$

We will separate the denominator factors and give each numerator a symbolic label, like $\text{\hspace{0.17em}}A,B\text{\hspace{0.17em},}$ or $\text{\hspace{0.17em}}C.$

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(x-1\right)}$

Multiply both sides of the equation by the common denominator to eliminate the fractions:

$\left(x+2\right)\left(x-1\right)\left[\frac{3x}{\left(x+2\right)\left(x-1\right)}\right]=\overline{)\left(x+2\right)}\left(x-1\right)\left[\frac{A}{\overline{)\left(x+2\right)}}\right]+\left(x+2\right)\overline{)\left(x-1\right)}\left[\frac{B}{\overline{)\left(x-1\right)}}\right]$

The resulting equation is

$3x=A\left(x-1\right)+B\left(x+2\right)$

Expand the right side of the equation and collect like terms.

$\begin{array}{l}3x=Ax-A+Bx+2B\\ 3x=\left(A+B\right)x-A+2B\end{array}$

Set up a system of equations associating corresponding coefficients.

$\begin{array}{l}3=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A+B\\ 0=-A+2B\end{array}$

Add the two equations and solve for $\text{\hspace{0.17em}}B.$

$\begin{array}{l}\underset{¯}{\begin{array}{l}3=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A+B\\ 0=-A+2B\end{array}}\\ 3\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}+3B\\ 1=B\end{array}$

Substitute $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ into one of the original equations in the system.

$\begin{array}{l}3=A+1\\ 2=A\end{array}$

Thus, the partial fraction decomposition is

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x-1\right)}$

Another method to use to solve for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ is by considering the equation that resulted from eliminating the fractions and substituting a value for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ that will make either the A - or B -term equal 0. If we let $\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}$ the
$A-$ term becomes 0 and we can simply solve for $\text{\hspace{0.17em}}B.$

Next, either substitute $\text{\hspace{0.17em}}B=1\text{\hspace{0.17em}}$ into the equation and solve for $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ or make the B -term 0 by substituting $\text{\hspace{0.17em}}x=-2\text{\hspace{0.17em}}$ into the equation.

We obtain the same values for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ using either method, so the decompositions are the same using either method.

$\frac{3x}{\left(x+2\right)\left(x-1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x-1\right)}$

Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the Heaviside method , named after Charles Heaviside, a pioneer in the study of electronics.

Find the partial fraction decomposition of the following expression.

$\frac{x}{\left(x-3\right)\left(x-2\right)}$

$\frac{3}{x-3}-\frac{2}{x-2}$

## Decomposing $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ Where Q(x) Has repeated linear factors

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.

## Partial fraction decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)\text{\hspace{0.17em}}$ Has repeated linear factors

The partial fraction decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)},\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}Q\left(x\right)\text{\hspace{0.17em}}$ has a repeated linear factor occurring $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ times and the degree of $\text{\hspace{0.17em}}P\left(x\right)\text{\hspace{0.17em}}$ is less than the degree of $\text{\hspace{0.17em}}Q\left(x\right),\text{\hspace{0.17em}}$ is

$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\frac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}$

Write the denominator powers in increasing order.

#### Questions & Answers

The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
Jhon Reply
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris

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