1 / 12

PAR TIAL FRAC TION

PAR TIAL FRAC TION. +. DECOMPOSITION. Let’s add the two fractions below. We need a common denominator:. In this section we are going to learn how to take this answer and “decompose” it meaning break down into the fractions that were added together to get this answer.

kuniko
Download Presentation

PAR TIAL FRAC TION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PAR TIALFRAC TION + DECOMPOSITION

  2. Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take this answer and “decompose” it meaning break down into the fractions that were added together to get this answer

  3. We start by factoring the denominator. There could have been a fraction for each factor of the denominator but we don’t know the numerators so we’ll call them A and B. (x+3)(x-2) (x+3)(x-2) (x+3)(x-2) Now we’ll clear this equation of fractions by multiplying every term by the common denominator.

  4. “The Convenient x Method for Solving” This equation needs to be true for any value of x. Let x = -3 B(0) = 0 We pick an x that will “conveniently” get rid of one of the variables and solve for the other.

  5. Now we’ll “conveniently” choose x to be 2 to get rid of A and find B. Let x = 2 A(0) = 0 4 3

  6. Summary of Partial Fraction Decomposition When Denominator Factors Into Linear Factors (Factors of first degree) Next we’ll look at repeated factors and quadratic factors Use “convenient” x method to find A, B, etc. Clear equation of fractions Set fraction equal to sum of fractions with each factor as a denominator using A, B, etc. for numerators Factor the denominator

  7. Partial Fraction Decomposition With Repeated Linear Factors When the denominator has a repeated linear factor, you need a fraction with a denominator for each power of the factor. (x-1)(x+2)2 (x-1)(x+2)2 (x-1)(x+2)2 (x-1)(x+2)2 Let x = 1

  8. 13 2 3 -2 Let x = -2 To find B we put A and C in and choose x to be any other number. Let x = 0

  9. Partial Fraction Decomposition With Quadratic Factors When the denominator has a quadratic factor (that won’t factor), you need a fraction with a linear numerator. (x+1)(x2+4) (x+1)(x2+4) (x+1)(x2+4) The convenient x method doesn’t work as nicely on these kind so we’ll use the “equating coefficients” method. First multiply everything out.

  10. No xterms on left No x2terms on left A = - B Look at x2 terms: 0 = A + B 0 = B + C C = - B Look at xterms: Look atterms with no x’s: 1 = 4A + C 1 = 4(-B) + (-B) Solve these. Substitution would probably be easiest. Look at the kinds of terms on each side and equate coefficients (meaning put the coefficients = to each other)

  11. Partial Fraction Decomposition With Repeated Quadratic Factors When the denominator has a repeated quadratic factor (that won’t factor), you need a fraction with a linear numerator for each power. (x2+4)2 (x2+4)2 (x2+4)2 multiply out equate coefficients of various kinds of terms (next screen)

  12. 1 -4 1 -4 1 1 Look at x3 terms: 1 = A 1 = B Look at x2 terms: Look at xterms: 0 = 4A+C 0 = 4(1)+C -4 =C Look at terms with no x: 0 = 4(1)+D -4 = D 0 = 4B+D

More Related