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# 9-4. Partial Fractions

Partial Fractions
In this section, you will:
• Decompose   P( x ) Q( x ) ,  where  Q( x )  has only nonrepeated linear factors.
• Decompose   P( x ) Q( x ) ,  where  Q( x )  has repeated linear factors.
• Decompose   P( x ) Q( x ) ,  where  Q( x )  has a nonrepeated irreducible quadratic factor.
• Decompose   P( x ) Q( x ) ,  where  Q( x )  has a repeated irreducible quadratic factor.

Earlier in this chapter, we studied systems of two equations in two variables, systems of three equations in three variables, and nonlinear systems. Here we introduce another way that systems of equations can be utilized—the decomposition of rational expressions.

Fractions can be complicated; adding a variable in the denominator makes them even more so. The methods studied in this section will help simplify the concept of a rational expression.

# Decomposing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  Where Q(x) Has Only Nonrepeated Linear Factors

Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at partial fraction decomposition, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified rational expression to the original expressions, called the partial fractions.

For example, suppose we add the following fractions:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>2</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−3 + −1 x+2

We would first need to find a common denominator,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mi>x</mi><mn>−3</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

Next, we would write each expression with this common denominator and find the sum of the terms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mfrac><mn>2</mn></mfrac></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x−3 ( x+2 x+2 )+ −1 x+2 ( x−3 x−3 )=                        2x+4−x+3 (x+2)(x−3) = x+7 x 2 −x−6

Partial fraction decomposition is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><munder><mrow><mfrac><mrow><mi>x</mi><mo>+</mo><mn>7</mn></mrow></mfrac></mrow></munder></mrow></annotation-xml></semantics>[/itex] x 2 −x−6 Simplified sum      = 2 x−3 + −1 x+2 Partial fraction decomposition

We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.

When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] x 2 −x−6 are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x−3 )( x+2 ), the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.

Partial Fraction Decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) :Q(x) Has Nonrepeated Linear Factors

The partial fraction decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]has nonrepeated linear factors and the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is less than the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q( x ) = A 1 ( a 1 x+ b 1 ) + A 2 ( a 2 x+ b 2 ) + A 3 ( a 3 x+ b 3 ) +⋅⋅⋅+ A n ( a n x+ b n ) .

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

1. Use a variable for the original numerators, usually<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mtext>, </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] A n  for each numerator
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q(x) = A 1 ( a 1 x+ b 1 ) + A 2 ( a 2 x+ b 2 ) +⋯+ A n ( a n x+ b n )
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.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+2 )( x−1 )

We will separate the denominator factors and give each numerator a symbolic label, like<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mtext> ,</mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+2 )( x−1 ) = A ( x+2 ) + B ( x−1 )

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x+2 )( x−1 )[ 3x ( x+2 )( x−1 ) ]= ( x+2 ) ( x−1 )[ A ( x+2 ) ]+( x+2 ) ( x−1 ) [ B ( x−1 ) ]

The resulting equation is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mn>3</mn><mi>x</mi><mo>=</mo><mi>A</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x−1 )+B( x+2 )

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mn>3</mn><mi>x</mi><mo>=</mo><mi>A</mi><mi>x</mi><mo>−</mo><mi>A</mi><mo>+</mo><mi>B</mi><mi>x</mi><mo>+</mo><mn>2</mn><mi>B</mi></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 3x=(A+B)x−A+2B

Set up a system of equations associating corresponding coefficients.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mn>3</mn><mo>=</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>A</mi><mo>+</mo><mi>B</mi></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 0=−A+2B

Add the two equations and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><munder accentunder="true"><mtable columnalign="left"><mtr><mtd><mn>3</mn><mo>=</mo><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mi>A</mi><mo>+</mo><mi>B</mi></mtd></mtr></mtable></munder></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 0=−A+2B ¯ 3 =     0  +3B 1=B

Substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into one of the original equations in the system.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mtable columnalign="left"><mtr><mtd><mn>3</mn><mo>=</mo><mi>A</mi><mo>+</mo><mn>1</mn></mtd></mtr></mtable></annotation-xml></semantics>[/itex] 2=A

Thus, the partial fraction decomposition is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+2 )( x−1 ) = 2 ( x+2 ) + 1 ( x−1 )

Another method to use to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]is by considering the equation that resulted from eliminating the fractions and substituting a value for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]that will make either the A- or B-term equal 0. If we let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]the

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>A</mi><mi>-</mi></mrow></annotation-xml></semantics>[/itex] term becomes 0 and we can simply solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>      </mtext><mn>3</mn><mi>x</mi><mo>=</mo><mi>A</mi><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mi>B</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]  3(1)=A[(1)−1]+B[(1)+2]         3=0+3B         1=B

Next, either substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the equation and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or make the B-term 0 by substituting<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>−2</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>         </mtext><mn>3</mn><mi>x</mi><mo>=</mo><mi>A</mi><mo stretchy="false">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mi>B</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]     3(−2)=A[(−2)−1]+B[(−2)+2]         −6=−3A+0          −6 −3 =A            2=A

We obtain the same values for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using either method, so the decompositions are the same using either method.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+2 )( x−1 ) = 2 ( x+2 ) + 1 ( x−1 )

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.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>x</mi></mfrac></mrow></annotation-xml></semantics>[/itex] ( x−3 )( x−2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>3</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−3 − 2 x−2

# Decomposing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) :Q(x) Has Repeated Linear Factors

The partial fraction decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) , when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]has a repeated linear factor occurring<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>n</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]times and the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is less than the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ), is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q( x ) = A 1 ( ax+b ) + A 2 ( ax+b ) 2 + A 3 ( ax+b ) 3 +⋅⋅⋅+ A n ( ax+b ) n

Write the denominator powers in increasing order.

Given a rational expression with repeated linear factors, decompose it.

1. Use a variable like<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the numerators and account for increasing powers of the denominators.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q(x) = A 1 (ax+b) + A 2 (ax+b) 2 + . . . +  A n (ax+b) n
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 with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mo>−</mo><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +2x+4 x 3 −4 x 2 +4x

The denominator factors are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><msup/></mrow></annotation-xml></semantics>[/itex] ( x−2 ) 2 . To allow for the repeated factor of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x−2 ), the decomposition will include three denominators:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x−2 ), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] ( x−2 ) 2 . Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mo>−</mo><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +2x+4 x 3 −4 x 2 +4x = A x + B ( x−2 ) + C ( x−2 ) 2

Next, we multiply both sides by the common denominator.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mi>x</mi><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (x−2) 2 [ − x 2 +2x+4 x (x−2) 2 ]=[ A x + B (x−2) + C (x−2) 2 ]x (x−2) 2                 − x 2 +2x+4=A (x−2) 2 +Bx(x−2)+Cx

On the right side of the equation, we expand and collect like terms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mo>−</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 +2x+4=A( x 2 −4x+4)+B( x 2 −2x)+Cx                       =A x 2 −4Ax+4A+B x 2 −2Bx+Cx                       =(A+B) x 2 +(−4A−2B+C)x+4A

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +2x+4=( A+B ) x 2 +( −4A−2B+C )x+4A
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mi>A</mi><mo>+</mo><mi>B</mi><mo>=</mo><mn>−1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (1) −4A−2B+C=2     (2) 4A=4     (3)

Solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi></mrow></annotation-xml></semantics>[/itex], we have

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>4</mn><mi>A</mi><mo>=</mo><mn>4</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]   A=1

Substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into equation (1).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>  </mtext><mi>A</mi><mo>+</mo><mi>B</mi><mo>=</mo><mn>−1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (1)+B=−1          B=−2

Then, to solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]substitute the values for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into equation (2).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mtext>      </mtext><mn>−4</mn><mi>A</mi><mn>−2</mn><mi>B</mi><mo>+</mo><mi>C</mi><mo>=</mo><mn>2</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] −4(1)−2(−2)+C=2             −4+4+C=2                            C=2

Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mo>−</mo><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +2x+4 x 3 −4 x 2 +4x = 1 x − 2 ( x−2 ) + 2 ( x−2 ) 2

Find the partial fraction decomposition of the expression with repeated linear factors.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>6</mn><mi>x</mi><mn>−11</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x−1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>6</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−1 − 5 ( x−1 ) 2

# Decomposing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) , Where Q(x) Has a Nonrepeated Irreducible Quadratic Factor

So far, we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]representing constants. Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>x</mi><mo>+</mo><mi>B</mi><mo>,</mo><mi>B</mi><mi>x</mi><mo>+</mo><mi>C</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]etc.

Decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) :Q(x) Has a Nonrepeated Irreducible Quadratic Factor

The partial fraction decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  such that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]has a nonrepeated irreducible quadratic factor and the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is less than the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is written as

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q( x ) = A 1 x+ B 1 ( a 1 x 2 + b 1 x+ c 1 ) + A 2 x+ B 2 ( a 2 x 2 + b 2 x+ c 2 ) +⋅⋅⋅+ A n x+ B n ( a n x 2 + b n x+ c n)

The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator:<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mo>,</mo><mi>C</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and so on.

Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.

1. Use variables such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the constant numerators over linear factors, and linear expressions such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] A 1 x+ B 1 , A 2 x+ B 2 , etc., for the numerators of each quadratic factor in the denominator.
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q(x) = A ax+b + A 1 x+ B 1 ( a 1 x 2 + b 1 x+ c 1 ) + A 2 x+ B 2 ( a 2 x 2 + b 2 x+ c 2 ) +⋅⋅⋅+ A n x+ B n ( a n x2 + b n x+ c n )
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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor

Find a partial fraction decomposition of the given expression.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>8</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20 ( x+3 )( x 2 +x+2 )

We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>8</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20 ( x+3 )( x 2 +x+2 ) = A ( x+3 ) + Bx+C ( x 2 +x+2 )

We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>3</mn><mo stretchy="false">)</mo><mo stretchy="false">(</mo><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 +x+2)[ 8 x 2 +12x−20 (x+3)( x 2 +x+2) ]=[ A (x+3) + Bx+C ( x 2 +x+2) ](x+3)( x 2 +x+2)                                       8 x 2 +12x−20=A( x 2 +x+2)+(Bx+C)(x+3)

Notice we could easily solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]by choosing a value for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]that will make the<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mi>x</mi><mo>+</mo><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]term equal 0. Let<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>x</mi><mo>=</mo><mn>−3</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and substitute it into the equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>              </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mn>8</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20=A( x 2 +x+2)+(Bx+C)(x+3)    8 (−3) 2 +12(−3)−20=A( (−3) 2 +(−3)+2)+(B(−3)+C)((−3)+3)                                         16=8A                                           A=2

Now that we know the value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]substitute it back into the equation. Then expand the right side and collect like terms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow/></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>8</mn><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20=2( x 2 +x+2)+(Bx+C)(x+3) 8 x 2 +12x−20=2 x 2 +2x+4+B x 2 +3B+Cx+3C 8 x 2 +12x−20=(2+B) x 2 +(2+3B+C)x+(4+3C)

Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>         </mtext><mn>2</mn><mo>+</mo><mi>B</mi><mo>=</mo><mn>8</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (1) 2+3B+C=12 (2)        4+3C=−20 (3)

Solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using equation (1) and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]using equation (3).

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>   </mtext><mn>2</mn><mo>+</mo><mi>B</mi><mo>=</mo><mn>8</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] (1)          B=6 4+3C=−20 (3)        3C=−24          C=−8

Thus, the partial fraction decomposition of the expression is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>8</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20 ( x+3 )( x 2 +x+2 ) = 2 ( x+3 ) + 6x−8 ( x 2 +x+2 )

Could we have just set up a system of equations to solve [link]?

Yes, we could have solved it by setting up a system of equations without solving for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]first. The expansion on the right would be:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mtable columnalign="left"><mtr><mtd><mrow/></mtd></mtr><mtr><mtd><mn>8</mn><msup/></mtd></mtr></mtable></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 2 +12x−20=A x 2 +Ax+2A+B x 2 +3B+Cx+3C 8 x 2 +12x−20=(A+B) x 2 +(A+3B+C)x+(2A+3C)

So the system of equations would be:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>         </mtext><mi>A</mi><mo>+</mo><mi>B</mi><mo>=</mo><mn>8</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] A+3B+C=12      2A+3C=−20

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −6x+7 ( x−1 )( x 2 +1 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>3</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−1 + 2x−4 x 2 +1

# Decomposing<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  When Q(x) Has a Repeated Irreducible Quadratic Factor

Now that we can decompose a simplified rational expression with an irreducible quadratic factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.

Decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  When Q(x) Has a Repeated Irreducible Quadratic Factor

The partial fraction decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) , when<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]has a repeated irreducible quadratic factor and the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>P</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ) is less than the degree of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] x ), is

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( a x 2 +bx+c ) n = A 1 x+ B 1 ( a x 2 +bx+c ) + A 2 x+ B 2 ( a x 2 +bx+c ) 2 + A 3 x+ B 3 ( a x 2 +bx+c ) 3 +⋅⋅⋅+ A nx+ B n ( a x 2 +bx+c ) n

Write the denominators in increasing powers.

Given a rational expression that has a repeated irreducible factor, decompose it.

1. Use variables like<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>,</mo><mi>B</mi><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]for the constant numerators over linear factors, and linear expressions such as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msub/></mrow></annotation-xml></semantics>[/itex] A 1 x+ B 1 , A 2 x+ B 2 , etc., for the numerators of each quadratic factor in the denominator written in increasing powers, such as
<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>P</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] Q(x) = A ax+b + A 1 x+ B 1 (a x 2 +bx+c) + A 2 x+ B 2 (a x 2 +bx+c) 2 +⋯+  A n + B n (a x 2 +bx+c) n
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 a Repeated Irreducible Quadratic Factor in the Denominator

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 4 + x 3 + x 2 −x+1 x ( x 2 +1 ) 2

The factors of the denominator are<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>,</mo><mo stretchy="false">(</mo><msup/></mrow></annotation-xml></semantics>[/itex] x 2 +1), and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><msup/></mrow></annotation-xml></semantics>[/itex] ( 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mi>x</mi><mo>+</mo><mi>B</mi><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]So, let’s begin the decomposition.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 4 + x 3 + x 2 −x+1 x ( x 2 +1 ) 2 = A x + Bx+C ( x 2 +1 ) + Dx+E ( x 2 +1 ) 2

We eliminate the denominators by multiplying each term by<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><msup/></mrow></annotation-xml></semantics>[/itex] ( x 2 +1 ) 2 . Thus,

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 4 + x 3 + x 2 −x+1=A ( x 2 +1 ) 2 +( Bx+C )( x )( x 2 +1 )+( Dx+E )(x)

Expand the right side.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mo> </mo><mtext>    </mtext><msup/></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex] x 4 + x 3 + x 2 −x+1=A( x 4 +2 x 2 +1)+B x 4 +B x 2 +C x 3 +Cx+D x 2 +Ex                                          =A x 4 +2A x 2+A+B x 4 +B x 2 +C x 3 +Cx+D x 2 +Ex

Now we will collect like terms.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><msup><mi>x</mi></msup></mrow></annotation-xml></semantics>[/itex] 4 + x 3 + x 2 −x+1=( A+B ) x 4 +( C ) x 3 +( 2A+B+D ) x 2 +( C+E )x+A

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mtext>         </mtext><mi>A</mi><mo>+</mo><mi>B</mi><mo>=</mo><mn>1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                 C=1 2A+B+D=1          C+E=−1                 A=1

We can use substitution from this point. Substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the first equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>1</mn><mo>+</mo><mi>B</mi><mo>=</mo><mn>1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]        B=0

Substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>=</mo><mn>0</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the third equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>2</mn><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mn>0</mn><mo>+</mo><mi>D</mi><mo>=</mo><mn>1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]                   D=−1

Substitute<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>C</mi><mo>=</mo><mn>1</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]into the fourth equation.

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtable columnalign="right"><mtr columnalign="right"><mtd columnalign="right"><mrow><mn>1</mn><mo>+</mo><mi>E</mi><mo>=</mo><mn>−1</mn></mrow></mtd></mtr></mtable></mrow></annotation-xml></semantics>[/itex]       E=−2

Now we have solved for all of the unknowns on the right side of the equal sign. We have<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>B</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>C</mi><mo>=</mo><mn>1</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mi>D</mi><mo>=</mo><mn>−1</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>E</mi><mo>=</mo><mn>−2.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We can write the decomposition as follows:

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 4 + x 3 + x 2 −x+1 x ( x 2 +1 ) 2 = 1 x + 1 ( x 2 +1 ) − x+2 ( x 2 +1 ) 2

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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 −4 x 2 +9x−5 ( x 2 −2x+3 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>x</mi><mn>−2</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −2x+3 + 2x+1 ( x 2 −2x+3 ) 2

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

# Key Concepts

• Decompose<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  by writing the partial fractions as<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  with repeated linear factors must account for the factors of the denominator in increasing powers. See [link].
• The decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x )  with a nonrepeated irreducible quadratic factor needs a linear numerator over the quadratic factor, as in<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] A x + Bx+C ( a x 2 +bx+c ) . See [link].
• In the decomposition of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] P( x ) Q( x ) , where<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>Q</mi><mrow><mo>(</mo></mrow></mrow></annotation-xml></semantics>[/itex] 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

<math display="block" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>A</mi><mi>x</mi><mo>+</mo><mi>B</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( a x 2 +bx+c ) + A 2 x+ B 2 ( a x 2 +bx+c ) 2 +⋯+ A n x+ B n ( a x 2 +bx+c ) n .

# 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,<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 x 2 +1  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<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mfrac/></mrow></annotation-xml></semantics>[/itex] 7x+13 3 x 2 +8x+15 = A x+1 + B 3x+5 , we eventually simplify to<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>7</mn><mi>x</mi><mo>+</mo><mn>13</mn><mo>=</mo><mi>A</mi><mo stretchy="false">(</mo><mn>3</mn><mi>x</mi><mo>+</mo><mn>5</mn><mo stretchy="false">)</mo><mo>+</mo><mi>B</mi><mo stretchy="false">(</mo><mi>x</mi><mo>+</mo><mn>1</mn><mo stretchy="false">)</mo><mo>.</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]Explain how you could intelligently choose an<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi></mrow></annotation-xml></semantics>[/itex]-value that will eliminate either<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]or<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and solve for<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]and<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>B</mi><mo>.</mo></mrow></annotation-xml></semantics>[/itex]

If we choose<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mn>−1</mn><mo>,</mo><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]then the B-term disappears, letting us immediately know that<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>A</mi><mo>=</mo><mn>3.</mn><mtext> </mtext></mrow></annotation-xml></semantics>[/itex]We could alternatively plug in<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mi>x</mi><mo>=</mo><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 5 3 , giving us a B-value of<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mtext> </mtext><mn>−2.</mn></mrow></annotation-xml></semantics>[/itex]

## Algebraic

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>16</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +10x+24

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi><mn>−79</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −5x−24

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>8</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x+3 − 5 x−8

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mo>−</mo><mi>x</mi><mn>−24</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −2x−24

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>10</mn><mi>x</mi><mo>+</mo><mn>47</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +7x+10

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x+5 + 9 x+2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>x</mi></mfrac></mrow></annotation-xml></semantics>[/itex] 6 x 2 +25x+25

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>32</mn><mi>x</mi><mn>−11</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 20 x 2 −13x+2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>3</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 5x−2 + 4 4x−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +7x+10

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −9

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>5</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 2( x+3 ) + 5 2( x−3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>10</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −25

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>6</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −4

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>3</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x+2 + 3 x−2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>x</mi><mn>−3</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −6x+5

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><mi>x</mi><mn>−1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −x−6

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>9</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 5( x+2 ) + 11 5( x−3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +8x+15

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><mi>x</mi><mn>−1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −5x+6

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>8</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−3 − 5 x−2

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−5</mn><mi>x</mi><mn>−19</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+4 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mi>x</mi></mfrac></mrow></annotation-xml></semantics>[/itex] ( x−2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 + 2 ( x−2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>7</mn><mi>x</mi><mo>+</mo><mn>14</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x+3 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−24</mn><mi>x</mi><mn>−27</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( 4x+5 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 6 4x+5 + 3 ( 4x+5 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−24</mn><mi>x</mi><mn>−27</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( 6x−7 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mo>−</mo><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x−7 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 x−7 − 2 ( x−7 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>14</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 x 2 +12x+18

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +20x+8 2x ( x+1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>4</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x − 3 2( x+1 ) + 7 2 ( x+1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +55x+25 5x ( 3x+5 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>54</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 3 +127 x 2 +80x+16 2 x 2 ( 3x+2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>4</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x + 2 x 2 − 3 3x+2 + 7 2 ( 3x+2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 −5 x 2 +12x+144 x 2 ( x 2 +12x+36 )

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +6x+11 ( x+2 )( x 2 +x+3 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +x+3 + 3 x+2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +9x+23 ( x−1 )( x 2 +6x+11 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−2</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +10x+4 ( x−1 )( x 2 +3x+8 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><mn>−3</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +3x+8 + 1 x−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 +3x+1 ( x+1 )( x 2 +5x−2 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +17x−1 ( x+3 )( x 2 +6x+1 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>x</mi><mn>−1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +6x+1 + 2 x+3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 ( x+5 )( x 2 +7x−5 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +5x+3 x 3 −1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +x+1 + 4 x−1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−5</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +18x−4 x 3 +8

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −7x+33 x 3 +27

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>2</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −3x+9 + 3 x+3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 +2x+40 x 3 −125

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>4</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +4x+12 8 x 3 −27

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 1 4 x 2 +6x+9 + 1 2x−3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−50</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +5x−3 125 x 3 −1

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>−2</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 3 −30 x 2 +36x+216 x 4 +216x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x + 1 x+6 − 4x x 2 −6x+36

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>3</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 3 +2 x 2 +14x+15 ( x 2 +4 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 +6 x 2 +5x+9 ( x 2 +1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>x</mi><mo>+</mo><mn>6</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +1 + 4x+3 ( x 2 +1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 − x 2 +x−1 ( x 2 −3 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 +5x+5 ( x+2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x+2 + 2x+3 ( x+2 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 +2 x 2 +4x ( x 2 +2x+9 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 +25 ( x 2 +3x+25 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 +3x+25 − 3x ( x 2 +3x+25 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 3 +11x+7x+70 ( 2 x 2 +x+14 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><mi>x</mi><mo>+</mo><mn>2</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x ( x 2 +4 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] 8x − x 8( x 2 +4 ) + 10−x 2 ( x 2 +4 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 4 + x 3 +8 x 2 +6x+36 x ( x 2 +6 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>x</mi><mn>−9</mn></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] ( x 2 −x ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 16 x − 9 x 2 + 16 x−1 − 7 ( x−1 ) 2

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>5</mn><msup/></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 3 −2x+1 ( x 2 +2x ) 2

## Extensions

For the following exercises, find the partial fraction expansion.

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 2 +4 ( x+1 ) 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x+1 − 2 ( x+1 ) 2 + 5 ( x+1 ) 3

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><msup><mi>x</mi></msup></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] 3 −4 x 2 +5x+4 ( x−2 ) 3

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

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>7</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x+8 + 5 x−2 − x−1 x 2 −6x−16

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>5</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−2 − 3 10( x+2 ) + 7 x+8 − 7 10( x−8 )

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mn>1</mn></mfrac></mrow></annotation-xml></semantics>[/itex] x−4 − 3 x+6 − 2x+7 x 2 +2x−24

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mfrac><mrow><mn>2</mn><mi>x</mi></mrow></mfrac></mrow></annotation-xml></semantics>[/itex] x 2 −16 − 1−2x x 2 +6x+8 − x−5 x 2 −4x

<math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><annotation-xml encoding="MathML-Content"><mrow><mo>−</mo><mfrac/></mrow></annotation-xml></semantics>[/itex] 5 4x − 5 2( x+2 ) + 11 2( x+4 ) + 5 4( x+4 )

## Glossary

partial fractions
the individual fractions that make up the sum or difference of a rational expression before combining them into a simplified rational expression
partial fraction decomposition
the process of returning a simplified rational expression to its original form, a sum or difference of simpler rational expressions