In this section we are going to discuss the process of decomposing a rational expression into sums and differences of simpler rational expressions. That is, we want to know what terms are added to yield a certain fraction. This process is called partial fraction decomposition.
Writing a fraction in terms of two or more TERMS is sometimes useful in calculus. While we could wait until your calculus class to discuss this topic, a possibility which just might find widespread support among college algebra students, caluclus applications are focused more on using this process as a step in a larger application than on teaching the process itself. For that reason we are introducing it here.
SAMPLE: Here is a typical question:
We are looking for the following:
The resulting terms are called the 'partial fraction decomposition' since each term independently forms just a part of the fraction.
Each of the examples will follow the same four steps.
FOUR STEPS TO PARTIAL DECOMPOSITIONCHECK: Is the fraction reduced? If not, divide out common factors.
Of the four steps to partial decomposition, Step 3 requires the most time. There are two methods that we will talk about that help us to solve for the unknowns. The two methods are called 'key number' method Examples 1 & 2, below) and 'comparison of coefficient method." In some problems, it is most effective to use a combination of the two methods (Example 3, below). EXAMPLE 1: Find the partial fraction decomposition of
CHECK: Is the fraction reduced? Yes. There are no common factors between the numerator and the denominator and the degree of the numerator is less than the degree of the denominator. The values x = 5 and x = - 3 are called key numbers. STEP 3: Use the key numbers to solve for the unknowns that will be easy to find.
EXAMPLE 2: Find the partial fraction decomposition of
CHECK: Is the fraction reduced? Yes. There are no common factors between the numerator and the denominator and the degree of the numerator is less than the degree of the denominator.
(2)2 + 2 ( 2) + 7 = A ( 2 - 1)2 + B ( 2) ( 2 - 1) + C (2) 15 = A + 2B + 2C But, we already found that C = 10 and A = 7, so 15 = 7 + 2B + 2(10) B = - 6 So, the partial fraction decomposition of the given fraction is written as follows:
Comparison of CoefficientsIn some partial decomposition problems, the number of unknowns (A, B, C, etc.) is more than the number of critical values (or key numbers). In that case we can combine the key number method with the method called comparison of coefficients to find all of the unknowns. Here is what we mean by 'comparison of coefficients'.
If the quadratic on the left is equal to the one on the right then 5 = a, 6 = b, and 9 = c.
EXAMPLE 3: Find the partial fraction decomposition of
STEP 3: The only key number is x = - 5. Note, there are no REAL numbers that will make (x2 + 1) equal to zero so it gives us no key numbers. 3 (-5)2 + 2(-5) - 1 = A (25 + 1) + (-5 B + C) (-5 + 5) 64 = 26 A A = (64/26) = (32/13) Since we still have two unknown values to find, we will turn to the comparison of coefficient method. First, we rearrange the identity in STEP 2 by grouping like terms together. Since we already know that A = (32/13) we can use A + B = 3 to find that B = (7/13) 5(7/13) + (- 9/13) = (26/13) = 2
The partial fraction decomposition is then written as follows:
© 1999 Jo Steig
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STEP 1: Write the partial fraction decomposition template (see text for an explanation).
