# 9.4 Partial fractions  (Page 5/7)

 Page 5 / 7

## Decomposing a rational function with a repeated irreducible quadratic factor in the denominator

Decompose the given expression that has a repeated irreducible factor in the denominator.

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}$

The factors of the denominator are $\text{\hspace{0.17em}}x,\left({x}^{2}+1\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\left({x}^{2}+1\right)}^{2}.\text{\hspace{0.17em}}$ Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form $\text{\hspace{0.17em}}Ax+B.\text{\hspace{0.17em}}$ So, let’s begin the decomposition.

$\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{A}{x}+\frac{Bx+C}{\left({x}^{2}+1\right)}+\frac{Dx+E}{{\left({x}^{2}+1\right)}^{2}}$

We eliminate the denominators by multiplying each term by $\text{\hspace{0.17em}}x{\left({x}^{2}+1\right)}^{2}.\text{\hspace{0.17em}}$ Thus,

${x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\left({x}^{2}+1\right)}^{2}+\left(Bx+C\right)\left(x\right)\left({x}^{2}+1\right)+\left(Dx+E\right)\left(x\right)$

Expand the right side.

Now we will collect like terms.

${x}^{4}+{x}^{3}+{x}^{2}-x+1=\left(A+B\right){x}^{4}+\left(C\right){x}^{3}+\left(2A+B+D\right){x}^{2}+\left(C+E\right)x+A$

Set up the system of equations matching corresponding coefficients on each side of the equal sign.

We can use substitution from this point. Substitute $\text{\hspace{0.17em}}A=1\text{\hspace{0.17em}}$ into the first equation.

Substitute $\text{\hspace{0.17em}}A=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B=0\text{\hspace{0.17em}}$ into the third equation.

Substitute $\text{\hspace{0.17em}}C=1\text{\hspace{0.17em}}$ into the fourth equation.

Now we have solved for all of the unknowns on the right side of the equal sign. We have $\text{\hspace{0.17em}}A=1,\text{\hspace{0.17em}}$ $B=0,\text{\hspace{0.17em}}$ $C=1,\text{\hspace{0.17em}}$ $D=-1,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}E=-2.\text{\hspace{0.17em}}$ We can write the decomposition as follows:

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

Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.

$\frac{{x}^{3}-4{x}^{2}+9x-5}{{\left({x}^{2}-2x+3\right)}^{2}}$

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

Access these online resources for additional instruction and practice with partial fractions.

## Key concepts

• Decompose $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ by writing the partial fractions as $\text{\hspace{0.17em}}\frac{A}{{a}_{1}x+{b}_{1}}+\frac{B}{{a}_{2}x+{b}_{2}}.\text{\hspace{0.17em}}$ Solve by clearing the fractions, expanding the right side, collecting like terms, and setting corresponding coefficients equal to each other, then setting up and solving a system of equations. See [link] .
• The decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ with repeated linear factors must account for the factors of the denominator in increasing powers. See [link] .
• The decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)}\text{\hspace{0.17em}}$ with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in $\text{\hspace{0.17em}}\frac{A}{x}+\frac{Bx+C}{\left(a{x}^{2}+bx+c\right)}.\text{\hspace{0.17em}}$ See [link] .
• In the decomposition of $\text{\hspace{0.17em}}\frac{P\left(x\right)}{Q\left(x\right)},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}Q\left(x\right)\text{\hspace{0.17em}}$ has a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as
$\frac{Ax+B}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}.$
See [link] .

## Verbal

Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction

No, a quotient of polynomials can only be decomposed if the denominator can be factored. For example, $\text{\hspace{0.17em}}\frac{1}{{x}^{2}+1}\text{\hspace{0.17em}}$ cannot be decomposed because the denominator cannot be factored.

Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)

Can you explain how to verify a partial fraction decomposition graphically?

Graph both sides and ensure they are equal.

You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.

Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had $\text{\hspace{0.17em}}\frac{7x+13}{3{x}^{2}+8x+15}=\frac{A}{x+1}+\frac{B}{3x+5},\text{\hspace{0.17em}}$ we eventually simplify to $\text{\hspace{0.17em}}7x+13=A\left(3x+5\right)+B\left(x+1\right).\text{\hspace{0.17em}}$ Explain how you could intelligently choose an $\text{\hspace{0.17em}}x$ -value that will eliminate either $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B.$

If we choose $\text{\hspace{0.17em}}x=-1,\text{\hspace{0.17em}}$ then the B -term disappears, letting us immediately know that $\text{\hspace{0.17em}}A=3.\text{\hspace{0.17em}}$ We could alternatively plug in $\text{\hspace{0.17em}}x=-\frac{5}{3},\text{\hspace{0.17em}}$ giving us a B -value of $\text{\hspace{0.17em}}-2.$

## Algebraic

For the following exercises, find the decomposition of the partial fraction for the nonrepeating linear factors.

$\frac{5x+16}{{x}^{2}+10x+24}$

$\frac{3x-79}{{x}^{2}-5x-24}$

$\frac{8}{x+3}-\frac{5}{x-8}$

$\frac{-x-24}{{x}^{2}-2x-24}$

$\frac{10x+47}{{x}^{2}+7x+10}$

$\frac{1}{x+5}+\frac{9}{x+2}$

$\frac{x}{6{x}^{2}+25x+25}$

$\frac{32x-11}{20{x}^{2}-13x+2}$

$\frac{3}{5x-2}+\frac{4}{4x-1}$

$\frac{x+1}{{x}^{2}+7x+10}$

$\frac{5x}{{x}^{2}-9}$

$\frac{5}{2\left(x+3\right)}+\frac{5}{2\left(x-3\right)}$

$\frac{10x}{{x}^{2}-25}$

$\frac{6x}{{x}^{2}-4}$

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

$\frac{2x-3}{{x}^{2}-6x+5}$

$\frac{4x-1}{{x}^{2}-x-6}$

$\frac{9}{5\left(x+2\right)}+\frac{11}{5\left(x-3\right)}$

$\frac{4x+3}{{x}^{2}+8x+15}$

$\frac{3x-1}{{x}^{2}-5x+6}$

$\frac{8}{x-3}-\frac{5}{x-2}$

For the following exercises, find the decomposition of the partial fraction for the repeating linear factors.

$\frac{-5x-19}{{\left(x+4\right)}^{2}}$

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

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

$\frac{7x+14}{{\left(x+3\right)}^{2}}$

$\frac{-24x-27}{{\left(4x+5\right)}^{2}}$

$-\frac{6}{4x+5}+\frac{3}{{\left(4x+5\right)}^{2}}$

$\frac{-24x-27}{{\left(6x-7\right)}^{2}}$

$\frac{5-x}{{\left(x-7\right)}^{2}}$

$-\frac{1}{x-7}-\frac{2}{{\left(x-7\right)}^{2}}$

$\frac{5x+14}{2{x}^{2}+12x+18}$

$\frac{5{x}^{2}+20x+8}{2x{\left(x+1\right)}^{2}}$

$\frac{4}{x}-\frac{3}{2\left(x+1\right)}+\frac{7}{2{\left(x+1\right)}^{2}}$

$\frac{4{x}^{2}+55x+25}{5x{\left(3x+5\right)}^{2}}$

$\frac{54{x}^{3}+127{x}^{2}+80x+16}{2{x}^{2}{\left(3x+2\right)}^{2}}$

$\frac{4}{x}+\frac{2}{{x}^{2}}-\frac{3}{3x+2}+\frac{7}{2{\left(3x+2\right)}^{2}}$

$\frac{{x}^{3}-5{x}^{2}+12x+144}{{x}^{2}\left({x}^{2}+12x+36\right)}$

For the following exercises, find the decomposition of the partial fraction for the irreducible nonrepeating quadratic factor.

$\frac{4{x}^{2}+6x+11}{\left(x+2\right)\left({x}^{2}+x+3\right)}$

$\frac{x+1}{{x}^{2}+x+3}+\frac{3}{x+2}$

$\frac{4{x}^{2}+9x+23}{\left(x-1\right)\left({x}^{2}+6x+11\right)}$

$\frac{-2{x}^{2}+10x+4}{\left(x-1\right)\left({x}^{2}+3x+8\right)}$

$\frac{4-3x}{{x}^{2}+3x+8}+\frac{1}{x-1}$

$\frac{{x}^{2}+3x+1}{\left(x+1\right)\left({x}^{2}+5x-2\right)}$

$\frac{4{x}^{2}+17x-1}{\left(x+3\right)\left({x}^{2}+6x+1\right)}$

$\frac{2x-1}{{x}^{2}+6x+1}+\frac{2}{x+3}$

$\frac{4{x}^{2}}{\left(x+5\right)\left({x}^{2}+7x-5\right)}$

$\frac{4{x}^{2}+5x+3}{{x}^{3}-1}$

$\frac{1}{{x}^{2}+x+1}+\frac{4}{x-1}$

$\frac{-5{x}^{2}+18x-4}{{x}^{3}+8}$

$\frac{3{x}^{2}-7x+33}{{x}^{3}+27}$

$\frac{2}{{x}^{2}-3x+9}+\frac{3}{x+3}$

$\frac{{x}^{2}+2x+40}{{x}^{3}-125}$

$\frac{4{x}^{2}+4x+12}{8{x}^{3}-27}$

$-\frac{1}{4{x}^{2}+6x+9}+\frac{1}{2x-3}$

$\frac{-50{x}^{2}+5x-3}{125{x}^{3}-1}$

$\frac{-2{x}^{3}-30{x}^{2}+36x+216}{{x}^{4}+216x}$

$\frac{1}{x}+\frac{1}{x+6}-\frac{4x}{{x}^{2}-6x+36}$

For the following exercises, find the decomposition of the partial fraction for the irreducible repeating quadratic factor.

$\frac{3{x}^{3}+2{x}^{2}+14x+15}{{\left({x}^{2}+4\right)}^{2}}$

$\frac{{x}^{3}+6{x}^{2}+5x+9}{{\left({x}^{2}+1\right)}^{2}}$

$\frac{x+6}{{x}^{2}+1}+\frac{4x+3}{{\left({x}^{2}+1\right)}^{2}}$

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

$\frac{{x}^{2}+5x+5}{{\left(x+2\right)}^{2}}$

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

$\frac{{x}^{3}+2{x}^{2}+4x}{{\left({x}^{2}+2x+9\right)}^{2}}$

$\frac{{x}^{2}+25}{{\left({x}^{2}+3x+25\right)}^{2}}$

$\frac{1}{{x}^{2}+3x+25}-\frac{3x}{{\left({x}^{2}+3x+25\right)}^{2}}$

$\frac{2{x}^{3}+11x+7x+70}{{\left(2{x}^{2}+x+14\right)}^{2}}$

$\frac{5x+2}{x{\left({x}^{2}+4\right)}^{2}}$

$\frac{1}{8x}-\frac{x}{8\left({x}^{2}+4\right)}+\frac{10-x}{2{\left({x}^{2}+4\right)}^{2}}$

$\frac{{x}^{4}+{x}^{3}+8{x}^{2}+6x+36}{x{\left({x}^{2}+6\right)}^{2}}$

$\frac{2x-9}{{\left({x}^{2}-x\right)}^{2}}$

$-\frac{16}{x}-\frac{9}{{x}^{2}}+\frac{16}{x-1}-\frac{7}{{\left(x-1\right)}^{2}}$

$\frac{5{x}^{3}-2x+1}{{\left({x}^{2}+2x\right)}^{2}}$

## Extensions

For the following exercises, find the partial fraction expansion.

$\frac{{x}^{2}+4}{{\left(x+1\right)}^{3}}$

$\frac{1}{x+1}-\frac{2}{{\left(x+1\right)}^{2}}+\frac{5}{{\left(x+1\right)}^{3}}$

$\frac{{x}^{3}-4{x}^{2}+5x+4}{{\left(x-2\right)}^{3}}$

For the following exercises, perform the operation and then find the partial fraction decomposition.

$\frac{7}{x+8}+\frac{5}{x-2}-\frac{x-1}{{x}^{2}-6x-16}$

$\frac{5}{x-2}-\frac{3}{10\left(x+2\right)}+\frac{7}{x+8}-\frac{7}{10\left(x-8\right)}$

$\frac{1}{x-4}-\frac{3}{x+6}-\frac{2x+7}{{x}^{2}+2x-24}$

$\frac{2x}{{x}^{2}-16}-\frac{1-2x}{{x}^{2}+6x+8}-\frac{x-5}{{x}^{2}-4x}$

$-\frac{5}{4x}-\frac{5}{2\left(x+2\right)}+\frac{11}{2\left(x+4\right)}+\frac{5}{4\left(x+4\right)}$

#### Questions & Answers

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
Carlos Reply
How can you tell what type of parent function a graph is ?
Mary Reply
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
Karim Reply
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
unknown Reply
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
Ef Reply
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
KARMEL Reply
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
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert

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