# 7.4 Partial fractions  (Page 5/7)

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## 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
$\frac{Ax+B}{\left(a{x}^{2}+bx+c\right)}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots \text{+}\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}.$

## 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.

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

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

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

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

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

An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×