# Partial Fractions Lesson 7.3, page 737 - PowerPoint PPT Presentation

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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
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

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.