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Partial Fractions Lesson 7.3, page 737

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Partial Fractions Lesson 7.3, page 737 Objective: To decompose rational expressions into partial fractions. Review: Rational Expressions rational function a ... – PowerPoint PPT presentation

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Title: Partial Fractions Lesson 7.3, page 737


1
Partial FractionsLesson 7.3, page 737
  • Objective To decompose rational expressions
    into partial fractions.

2
Review Rational Expressions
  • rational function a quotient of two polynomials
  • where p(x) and q(x) are polynomials and where
    q(x) is not the zero polynomial.

3
ReviewAddingSubtracting Rationals
4
Definition
In the problem above, the fractions on the right
are called partial fractions. In calculus, it
is often helpful to write a rational expression
as the sum of two or more simpler rational
expressions.
5
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!

6
Example 1
  • Decompose into partial fractions.
  • 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 add the expressions
    on the rightgiving us

7
  • Equate the numerators 4x ? 13 A(2x ? 3) B(x
    2)
  • Since the above 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.
  • B ?2.

8
Example 1 continued
  • If we choose x ?2, then x 2 0 and B will be
    eliminated when we make the substitution. So,
    4(?2) ? 13 A2(?2) ? 3 B(?2 2)
  • ?21 ?7A.
  • A 3.
  • The decomposition is as follows


9
Check Point 1, page 740Find the partial fraction
decomposition.
10
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 if one denominator is a linear term cubed?
    There would be 3 denominators in the
    decomposition

11
Example 2 Decompose into partial fractions.
12
Example 2 continuedNext, we add the expression
on the right
  • Then, we equate the numerators. 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 ½ . This gives us

Solving, we obtain A 5.
13
Example 2 continuedIn order to have x ? 2 0,
we let x 2.
  • Substituting gives us
  • To find B, we choose any value for x except ½ or
    2
  • and replace A with 5 and C with ?2 . We let x
    1

14
Example 2 continued The decomposition is as
follows
15
Check Point 2, page 741Find the partial fraction
decomposition.
16
Check Point 3, page 743Find the partial fraction
decomposition.
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