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

  1. Use a variable like A , B , or C for the numerators and account for increasing powers of the denominators.
    P ( x ) Q ( x ) = A 1 ( a x + b ) + A 2 ( a x + b ) 2 +   . +  A n ( a 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 with repeated linear factors

Decompose the given rational expression with repeated linear factors.

x 2 + 2 x + 4 x 3 −4 x 2 + 4 x

The denominator factors are x ( x −2 ) 2 . To allow for the repeated factor of ( x −2 ) , the decomposition will include three denominators: x , ( x −2 ) , and ( x −2 ) 2 . Thus,

x 2 + 2 x + 4 x 3 −4 x 2 + 4 x = A x + B ( x −2 ) + C ( x −2 ) 2

Next, we multiply both sides by the common denominator.

x ( x −2 ) 2 [ x 2 + 2 x + 4 x ( x −2 ) 2 ] = [ A x + B ( x −2 ) + C ( x −2 ) 2 ] x ( x −2 ) 2                  x 2 + 2 x + 4 = A ( x −2 ) 2 + B x ( x −2 ) + C x

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

x 2 + 2 x + 4 = A ( x 2 4 x + 4 ) + B ( x 2 2 x ) + C x                        = A x 2 4 A x + 4 A + B x 2 2 B x + C x                        = ( A + B ) x 2 + ( 4 A 2 B + C ) x + 4 A

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

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

Solving for A , we have

4 A = 4    A = 1

Substitute A = 1 into equation (1).

   A + B = −1 ( 1 ) + B = −1           B = −2

Then, to solve for C , substitute the values for A and B into equation (2).

       −4 A −2 B + C = 2 −4 ( 1 ) −2 ( −2 ) + C = 2              −4 + 4 + C = 2                             C = 2

Thus,

x 2 + 2 x + 4 x 3 −4 x 2 + 4 x = 1 x 2 ( x −2 ) + 2 ( x −2 ) 2
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Find the partial fraction decomposition of the expression with repeated linear factors.

6 x −11 ( x −1 ) 2

6 x −1 5 ( x −1 ) 2

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Decomposing 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 A , B , or C 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 A x + B , B x + C , etc.

Decomposition of P ( x ) Q ( x ) : Q ( x ) Has a nonrepeated irreducible quadratic factor

The partial fraction decomposition of P ( x ) Q ( x ) such that Q ( x ) has a nonrepeated irreducible quadratic factor and the degree of P ( x ) is less than the degree of Q ( x ) is written as

P ( x ) 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: A , B , C , 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 A , B , or C for the constant numerators over linear factors, and linear expressions such as A 1 x + B 1 , A 2 x + B 2 , etc., for the numerators of each quadratic factor in the denominator.
    P ( x ) Q ( x ) = A a x + 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 x 2 + 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.

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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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