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Decomposition-of-Fractions

Page history last edited by Kristen Fouss 11 years, 7 months ago

 

Chapter 7.3 - Partial Fraction Decomposition  

 

 

 

    

OBJECTIVE

 

Fraction decomposition is used to take a fraction with a complex denominator and break it into smaller, linear denominator fractions.  This method was first introduced by John Bernoulli (1667-1748) http://en.wikipedia.org/wiki/Johann_Bernoulli.

 

 

STEP BY STEP METHOD

0.  (Only necessary if fraction is improper)  If eq=\frac{N(x)}{D(x)}  is an improper fraction (degree of N(x) is greater than D(x)) divide the two polynomials.  If the numerator is 1 degree higher than the denominator, synthetic division can be used.

 *Now take the eq=\frac{remainder}{D(x)} and use the following steps to break it down.

 

1.  Factor the denominator into linear factors and set each term as an individual quotient with variables A, B and C (if needed) as the denominators.

*If a linear term repeats, you must include all more simple forms. For instance x^2 must be represented with x and x^2

  

2.  Multiply each term by the LCD (lowest common denominator) to eliminate the fractions.

 

3. Substitute x with a number that causes the A term to drop out and solve to determine a value for B

 

4. Repeat step for except make B drop out.

 

5. Use the factored form, individual quotients from step 2 and plug in the values for A and B. If step 0 applied, add the results to this step.

 

ADDITIONAL STEPS IF USING A, B, and C

 

As stated in the above steps, if a linear term repeats, you must include all more simple forms. So if the denominator is   eq=x(x+1)^2 , you must include x, x +1, and eq=(x + 1)^2 .  When this occurs two terms may drop out instead of one.  To figure out what the terms are, you must instead compare the two sides of the equation.  If there is eq=Ax^2  on one side and  eq=5x^2 on the other (and no other squared terms), A must be equal to 5.  Use this method to figure out the other terms, pluging in the ones you've figured out to help.  Again, this is only necessary if more than one of the terms drop out.

 

 

DECOMPOSITION IN ACTION

decomposition for example 2

Multiply both side of the above equation by (x + 1)2, and simplify to obtain an equation of the form

1 - 2x = A(x + 1) + B

Expand the right side and group like terms

-2x + 1 = A x + (A + B)

Match up the variables with the polynomial in its place on the other side of the equals sign [(-2x) and (Ax), (A+B) and 1].For the right and left polynomials to be equal we need to have

- 2 = A and 1 = A + B

Solve the above system to obtain

A = - 2 and B = 3

 

solution for example 2

 

 

EXTRAS

 

For more information see this 20 minute partial fraction decomposition lesson/tutorial:  http://video.google.com/videoplay?docid=977064265128829595&ei=9IFGSdTlC4fIqAKL9pidDw&q=partial+fraction+decomposition&hl=en

 

Partial fraction decomposition generator.  Simply enter the fraction (use ^ for powers, * for multiplication, / for division) and click 'Partial Fractions'.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=partial_fractions&s3=basic

 

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