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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.

x 4 + x 3 + x 2 x + 1 x ( x 2 + 1 ) 2

The factors of the denominator are x , ( x 2 + 1 ) , and ( x 2 + 1 ) 2 . 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 A x + B . So, let’s begin the decomposition.

x 4 + x 3 + x 2 x + 1 x ( x 2 + 1 ) 2 = A x + B x + C ( x 2 + 1 ) + D x + E ( x 2 + 1 ) 2

We eliminate the denominators by multiplying each term by x ( x 2 + 1 ) 2 . Thus,

x 4 + x 3 + x 2 x + 1 = A ( x 2 + 1 ) 2 + ( B x + C ) ( x ) ( x 2 + 1 ) + ( D x + E ) ( x )

Expand the right side.

       x 4 + x 3 + x 2 x + 1 = A ( x 4 + 2 x 2 + 1 ) + B x 4 + B x 2 + C x 3 + C x + D x 2 + E x                                         = A x 4 + 2 A x 2 + A + B x 4 + B x 2 + C x 3 + C x + D x 2 + E x

Now we will collect like terms.

x 4 + x 3 + x 2 x + 1 = ( A + B ) x 4 + ( C ) x 3 + ( 2 A + B + D ) x 2 + ( C + E ) x + A

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

          A + B = 1                  C = 1 2 A + B + D = 1           C + E = −1                  A = 1

We can use substitution from this point. Substitute A = 1 into the first equation.

1 + B = 1         B = 0

Substitute A = 1 and B = 0 into the third equation.

2 ( 1 ) + 0 + D = 1                    D = −1

Substitute C = 1 into the fourth equation.

1 + E = −1        E = −2

Now we have solved for all of the unknowns on the right side of the equal sign. We have A = 1 , B = 0 , C = 1 , D = −1 , and E = −2. We can write the decomposition as follows:

x 4 + x 3 + x 2 x + 1 x ( x 2 + 1 ) 2 = 1 x + 1 ( x 2 + 1 ) x + 2 ( x 2 + 1 ) 2
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Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.

x 3 −4 x 2 + 9 x −5 ( x 2 −2 x + 3 ) 2

x −2 x 2 −2 x + 3 + 2 x + 1 ( x 2 −2 x + 3 ) 2

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

Key concepts

  • Decompose P ( x ) Q ( x ) by writing the partial fractions as A a 1 x + b 1 + B a 2 x + b 2 . 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 P ( x ) Q ( x ) with repeated linear factors must account for the factors of the denominator in increasing powers. See [link] .
  • The decomposition of P ( x ) Q ( x ) with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in A x + B x + C ( a x 2 + b x + c ) . See [link] .
  • In the decomposition of P ( x ) Q ( x ) , where Q ( x ) 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
    A x + B ( a x 2 + b x + c ) + A 2 x + B 2 ( a x 2 + b x + c ) 2 + + A n x + B n ( a x 2 + b x + c ) n .
    See [link] .

Section exercises

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, 1 x 2 + 1 cannot be decomposed because the denominator cannot be factored.

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

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

Graph both sides and ensure they are equal.

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

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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 7 x + 13 3 x 2 + 8 x + 15 = A x + 1 + B 3 x + 5 , we eventually simplify to 7 x + 13 = A ( 3 x + 5 ) + B ( x + 1 ) . Explain how you could intelligently choose an x -value that will eliminate either A or B and solve for A and B .

If we choose x = −1 , then the B -term disappears, letting us immediately know that A = 3. We could alternatively plug in x = 5 3 , giving us a B -value of −2.

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Algebraic

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

5 x + 16 x 2 + 10 x + 24

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3 x −79 x 2 −5 x −24

8 x + 3 5 x −8

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x −24 x 2 −2 x −24

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10 x + 47 x 2 + 7 x + 10

1 x + 5 + 9 x + 2

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32 x −11 20 x 2 −13 x + 2

3 5 x −2 + 4 4 x −1

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5 x x 2 −9

5 2 ( x + 3 ) + 5 2 ( x −3 )

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6 x x 2 −4

3 x + 2 + 3 x −2

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4 x −1 x 2 x −6

9 5 ( x + 2 ) + 11 5 ( x −3 )

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3 x −1 x 2 −5 x + 6

8 x −3 5 x −2

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

−5 x −19 ( x + 4 ) 2

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x ( x −2 ) 2

1 x −2 + 2 ( x −2 ) 2

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−24 x −27 ( 4 x + 5 ) 2

6 4 x + 5 + 3 ( 4 x + 5 ) 2

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−24 x −27 ( 6 x −7 ) 2

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5 x ( x −7 ) 2

1 x −7 2 ( x −7 ) 2

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5 x + 14 2 x 2 + 12 x + 18

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5 x 2 + 20 x + 8 2 x ( x + 1 ) 2

4 x 3 2 ( x + 1 ) + 7 2 ( x + 1 ) 2

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4 x 2 + 55 x + 25 5 x ( 3 x + 5 ) 2

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54 x 3 + 127 x 2 + 80 x + 16 2 x 2 ( 3 x + 2 ) 2

4 x + 2 x 2 3 3 x + 2 + 7 2 ( 3 x + 2 ) 2

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x 3 −5 x 2 + 12 x + 144 x 2 ( x 2 + 12 x + 36 )

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

4 x 2 + 6 x + 11 ( x + 2 ) ( x 2 + x + 3 )

x + 1 x 2 + x + 3 + 3 x + 2

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4 x 2 + 9 x + 23 ( x −1 ) ( x 2 + 6 x + 11 )

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−2 x 2 + 10 x + 4 ( x −1 ) ( x 2 + 3 x + 8 )

4 −3 x x 2 + 3 x + 8 + 1 x −1

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x 2 + 3 x + 1 ( x + 1 ) ( x 2 + 5 x −2 )

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4 x 2 + 17 x −1 ( x + 3 ) ( x 2 + 6 x + 1 )

2 x −1 x 2 + 6 x + 1 + 2 x + 3

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4 x 2 ( x + 5 ) ( x 2 + 7 x −5 )

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4 x 2 + 5 x + 3 x 3 −1

1 x 2 + x + 1 + 4 x −1

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−5 x 2 + 18 x −4 x 3 + 8

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3 x 2 −7 x + 33 x 3 + 27

2 x 2 −3 x + 9 + 3 x + 3

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x 2 + 2 x + 40 x 3 −125

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4 x 2 + 4 x + 12 8 x 3 −27

1 4 x 2 + 6 x + 9 + 1 2 x −3

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−50 x 2 + 5 x −3 125 x 3 −1

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−2 x 3 −30 x 2 + 36 x + 216 x 4 + 216 x

1 x + 1 x + 6 4 x x 2 −6 x + 36

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

3 x 3 + 2 x 2 + 14 x + 15 ( x 2 + 4 ) 2

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x 3 + 6 x 2 + 5 x + 9 ( x 2 + 1 ) 2

x + 6 x 2 + 1 + 4 x + 3 ( x 2 + 1 ) 2

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x 3 x 2 + x −1 ( x 2 −3 ) 2

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x 2 + 5 x + 5 ( x + 2 ) 2

x + 1 x + 2 + 2 x + 3 ( x + 2 ) 2

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x 3 + 2 x 2 + 4 x ( x 2 + 2 x + 9 ) 2

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x 2 + 25 ( x 2 + 3 x + 25 ) 2

1 x 2 + 3 x + 25 3 x ( x 2 + 3 x + 25 ) 2

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2 x 3 + 11 x + 7 x + 70 ( 2 x 2 + x + 14 ) 2

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5 x + 2 x ( x 2 + 4 ) 2

1 8 x x 8 ( x 2 + 4 ) + 10 x 2 ( x 2 + 4 ) 2

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x 4 + x 3 + 8 x 2 + 6 x + 36 x ( x 2 + 6 ) 2

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2 x −9 ( x 2 x ) 2

16 x 9 x 2 + 16 x −1 7 ( x −1 ) 2

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5 x 3 −2 x + 1 ( x 2 + 2 x ) 2

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Extensions

For the following exercises, find the partial fraction expansion.

x 2 + 4 ( x + 1 ) 3

1 x + 1 2 ( x + 1 ) 2 + 5 ( x + 1 ) 3

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x 3 −4 x 2 + 5 x + 4 ( x −2 ) 3

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

7 x + 8 + 5 x −2 x −1 x 2 −6 x −16

5 x −2 3 10 ( x + 2 ) + 7 x + 8 7 10 ( x −8 )

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1 x −4 3 x + 6 2 x + 7 x 2 + 2 x −24

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2 x x 2 −16 1 −2 x x 2 + 6 x + 8 x −5 x 2 −4 x

5 4 x 5 2 ( x + 2 ) + 11 2 ( x + 4 ) + 5 4 ( x + 4 )

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Questions & Answers

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-4
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x=-4
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x=-1
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Adityasuman x= - 1
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Practice Key Terms 2

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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