# 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.

how can are find the domain and range of a relations
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?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
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.
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?
what is foci?
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
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim