7.3 Partial Fractions

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7.3Partial Fractions

• Decompose rational expressions into partial
  fractions.

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What is decomposition of partial fractions?

• Writing a more complex fraction as the sum or
  difference of simpler fractions.
• Examples
• Why would you ever want to do this? Its
  EXTREMELY helpful in calculus!

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Partial Fractions
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Distinct linear factors

• Factor the denominator into linear terms
• Each linear term will be the denominator of a
  separate term (i.e. if there are 3 factors, there
  will be 3 separate fractions added together)
• Example next slide

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• Decompose into partial fractions
  .
• Solution The degree of the numerator is less
  than the degree of the denominator. We begin by
  factoring the denominator (x 2)(2x ? 3). We
  know that there are constants A and B such that
• 
  
• To determine A and B, we first multiply both
  sides of the equation by the LCD (x 2)(2x ? 3).
  

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• 
  
• The result is 4x ? 13 A(2x ? 3) B(x 2).
• Since the last equation containing A and B is
  true for all x, we can substitute any value of x
  and still have a true equation. If we choose x
  3/2, then 2x ? 3 0 and A will be eliminated
  when we make the substitution. This gives us
• 4(3/2) ? 13 A2(3/2) ? 3 B(3/2 2)
• ?7 0 (7/2)B.
• Solving we obtain B ?2.

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• If we choose x ?2, then x 2 0 and B will be
  eliminated when we make the substitution. This
  gives us
• 4(?2) ? 13 A2(?2) ? 3 B(?2 2)
• ?21 ?7A.
• Solving, we obtain A 3.
• The decomposition is as follows
• 
  .

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What if one on the denominators is a linear term
squared?

• This is accounted for by having the nonsquared
  term as one denominator and having the squared
  term as another denominator.
• What is one denominator is a linear term cubed?
  There would be 3 denominators in the
  decomposition

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• Decompose into partial fractions
• 
  .
• Solution The degree of the numerator is 2
  and the degree of the denominator is 3, so the
  degree of the numerator is less than the degree
  of the denominator. The denominator is given in
  factored form. The decomposition has the
  following form
• 
  .

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• Next, we multiply both sides by LCD
• 
  
• This gives us
• 
  
• Since the equation containing A, B, and C is true
  for all of x, we can substitute any value of x
  and still have a true equation. In order to have
  2x 1 0, we let x . This gives us

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• Solving, we obtain A 5.
• In order to have x ? 2 0, we let x 2.
  Substituting gives us
• Solving, we obtain C ?2 .
• To find B, we choose any value for x except or
  2 and replace A with 5 and C with ?2 . We let x
  1

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Another Example continued

• The decomposition is as follows

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What if denominator has a quadratic factor (not
reduced to product of linear factors)?

• A quadratic denominator in decomposition would
  have a linear numerator
• To decompose the fraction, you proceed precisely
  as was done with linear denominators.

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• Decompose into partial fractions
  .
• Solution The decomposition has the following
  form.
• Multiplying by the LCD, we get

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• We then equate corresponding coefficients
• 11 A 2C, The coefficients of the
  x2-terms
• ?8 ?3A B, The coefficients of the
  x-terms
• ?7 ?3B ? C. The constant terms
• We solve this system of three equations and
  obtain
• A 3, B 1, and C 4.
• The decomposition is as follows