Partial Fractions. by Richard Gill. Supported in part by funding from a VCCS LearningWare Grant

Transcription

1 Partial Fractions by Richard Gill Supported in part by funding from a VCCS LearningWare Grant

2 EXMPLE : For our first eample we will work an LCD problem frontwards and backwards. Use an LCD to complete the following addition. The LCD is ( + )( ). We now convert each fraction to LCD status. On the net slide we will work this problem backwards 7 0 7

3 7 Find the partial fraction decomposition for: s we saw in the previous slide the denominator factors as ( + )( ). We want to find numbers and so that: 7

4 Partial Fraction Decomposition Forms Distinct Linear Factors: 7 Repeated Linear Factors: 0 6 C

5 Partial Fraction Decomposition Forms C Repeated Quadratic Factors: Distinct Linear and Quadratic Factors: D C

6 7 Find the partial fraction decomposition for:

7 Now we epand and compare the left side to the right side If the left side and the right side are going to be equal then: + has to be and -+ has to be 7.

8 This gives us two equations in two unknowns. We can add the two equations and finish it off with back substitution. + = - + = 7 = = If = and + = then =. Cool!! ut what does this mean?

9 Remember that our original mission was to break a big fraction into a couple of pieces. In particular to find and so that: 7 We now know that = and = which means that 7 Now we will look at this same strategy applied to an LCD with one linear factor and one quadratic factor in the denominator. nd that is partial fraction decomposition!

10 EXMPLE : Find the partial fraction decomposition for 0 First we will see if the denominator factors. (If it doesn t we are doomed.) The denominator has four terms so we will try to factor by grouping.

11 Since the denominator is factorable we can pursue the decomposition. 0 C 0 C 0 C 0 C C C ( ) ( C)

12 From this point 0 0 C Use Judicious Substitution: Let = -: 0 C 9 0 C 9 C C Two reasons why using Judicious Substitution does not work for the other term: - It has no real zeros. - Even if it did, it still leaves two unknowns, &, that cannot be solved using a single equation. 0

13 Continuing on from this same point 0 Epand the right side completely: Collect like -terms: 0 terms: C C C C C ssociate like -terms and factor out : C C Now compare left- side terms with right-side terms: terms: constants: C ( ) ( C) C C 0

14 Note how easily this system of equations can be solved using the single value we were able to obtain from Judicious Substitution? Recall that C = -: C C 0 Substituting C into the first equation: Substituting C into the third equation: 0 6

15 Summarizing: C On the net slide, we solve this system of equations as if we did not know the value of C.

16 - = + C = + -0 = + C dd this equation to eliminate. Multiply both sides by - dd these two equations to eliminate. We now have two equations in and C. Compare the coefficients. = - - C = + = C Multiply both sides of this equation by. - = - + 9C -0 = + C - = C - = C - = + C - = - = We can finish by back substitution. -0 = + C -0 = + (-) = =

17 We have now discovered that =, = and C = -. Fair enough. We began with the idea that we could break the following fraction up into smaller pieces (partial fraction decomposition). 0 0 C OK, but I forgot what this means. Substitute for, and C and we are done.

18 EXMPLE : For our net eample, we are going to consider what happens when one of the factors in the denominator is raised to a power. Consider the following for partial fraction decomposition: There are two setups that we could use to begin: Setup proceeds along the same lines as the 7 previous eample. Setup considers that the second fraction could have come from two pieces. 7 7 C ( ) ( ) C

19 Since we have already done an eample with Setup, this eample will proceed with Setup. Step will be to multiply both sides by the LCD and simplify ( C C ) 6 C 9 C C C We now compare the coefficients of the two sides. Epand. Group like terms and factor.

20 The last line of the previous slide left us here. 6 C 7 9 If we compare the coefficients on each side, we have: + = C = 9 = 7 From the third equation =. Substituting into the first equation: + = so + = and =. Substituting back into the second equation: C = so 6() + () + C = + + C = 6 + C = and C = -

21 To refresh your memory, we were looking for values of of, and C that would satisfy the partial fraction decomposition below and we did find that =, = and C=-. 7 ( ) C So.. 7 ( ) Our last eample considers the possibility that the polynomial in the denominator has a smaller degree than the polynomial in the numerator.

22 EXMPLE : Find the partial fraction decomposition for Since the order of the numerator is larger than the order of the denominator, the first step is division. 6

23 y long division we have discovered that: We will now do partial fraction decomposition on the remainder.

24 Multiply both sides by the LCD. Distribute Group like terms Compare coefficients

25 From the previous slide we have that: If these two sides are equal then: = + and = To eliminate multiply both sides of the first equation by and add. = - = - -6 = so = -/ If + = and = -/ then / = / and = /

26 / / In summary then:

27 Tips for partial fraction decomposition of N()/D():. If N() has a larger order than D(), begin by long division. Then eamine the remainder for decomposition.. Factor D() into factors of (a + b) and a b c. If the factor (a + b) repeats then the decomposition must include:. If the factor a b a a b b c decomposition must include: repeats then the a b c C D a b c You should now check out the companion piece to this tutorial, which contains practice problems, their answers and several complete solutions.