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Section 6.3 Partial Fractions

Section 6.3 Partial Fractions. Advanced Algebra. What is Partial Fraction Decomposition?. There are times when we are working with Rational Functions of the form when we want to

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Section 6.3 Partial Fractions

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  1. Section 6.3 Partial Fractions Advanced Algebra

  2. What is Partial Fraction Decomposition? There are times when we are working with Rational Functions of the form when we want to split it up into two simpler fractions. The process that we go through to do this is called “partial fraction decomposition”

  3. Let’s say you had the following: What would you need to do to add them together? Multiply by a common denominator to both fractions Write as one fraction and simplify the numerator The final answer Partial Fraction Decomposition is the reverse of what we just did here…

  4. Break into 2 smaller fractions Multiply by LCD Cancel anything necessary Collect like terms and set equal

  5. Multiply by the common denominator. Collect like terms together and set them equal to each other. Solve two equations with two unknowns.

  6. This technique is called Partial Fractions Solve two equations with two unknowns.

  7. Sometimes you might get a repeated factor (multiplicity) in the denominator Repeated roots: we must use two terms for partial fractions.

  8. Partial-Fraction Decomposition Repeated linear factor

  9. Now we have 3 equations with 3 unknowns. Solve like in previous section.

  10. If the degree of the numerator is higher than the degree of the denominator, use long division first. (from example one)

  11. What if the denominator has a non-factorable quadratic in it? Partial Fraction Decomposition can get very complicated, very quickly when there are non-factorable quadratics and repeated linear factors…here is an easy example… Now just solve the 3 by 3 system… A=3, B=2 and C=4

  12. A nice shortcut if you have non-repeated linear factors—the Heaviside Shortcut—named after mathematician Oliver Heaviside (1850-1925)… Tell yourself, “if x is 5, then x-5 is 0.” Cover up the x-5 and put 5 in for the x in what is left… Do the same for the other linear factor Which should probably be simplified…

  13. A challenging example: first degree numerator irreducible quadratic factor repeated root

  14. We can do this problem on the TI-89: Of course with the TI-89, we could just integrate and wouldn’t need partial fractions! expand ((-2x+4)/((x^2+1)*(x-1)^2)) F2 3 p

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