Partial Fractions - PowerPoint PPT Presentation

Loading...

PPT – Partial Fractions PowerPoint presentation | free to view - id: 26ce6e-Nzg5Y



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Partial Fractions

Description:

Example 1. Decompose into partial fractions: ... If we choose x = 3/2, then 2x 3 = 0 and A will be eliminated when we make the substitution. ... – PowerPoint PPT presentation

Number of Views:161
Avg rating:3.0/5.0
Slides: 13
Provided by: tinae
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Partial Fractions


1
Section 5.8
  • Partial Fractions

2
Partial Fractions
3
Example 1
  • Decompose into partial fractions
  • Solution The degree of the numerator is less
    than the degree of the denominator.
  • 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 right

4
Example 1 continued
  • 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.

5
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


6
Example 2
  • 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

7
Example 2 continued
  • Next, 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.
8
Example 2 continued
  • In 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

9
Example 2 continued
  • The decomposition is as follows

10
Example 3
  • Decompose into partial fractions

11
Example 3 continued
  • Solution
  • The decomposition has
  • the following form.
  • Adding and equating numerators, we get
  • or

12
Example 3 continued
  • 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
About PowerShow.com