1. Partial Fractions

2. In Calculus, there are several procedures that are much easier if we can take a rather large fraction and break it up into pieces. The procedure that can decompose larger fractions is called Partial Fraction Decompostition. We will proceed as if we are working backwards through an addition of fractions with LCD. EXAMPLE 1: For our first example we will work an LCD problem frontwards and backwards. Use an LCD to complete the following addition. The LCD is (x + 2)(x – 1). We now convert each fraction to LCD status. On the next slide we will work this problem backwards

3. Find the partial fraction decomposition for: As we saw in the previous slide the denominator factors as (x + 2)(x – 1). We want to find numbers A and B so that: The bad news is that we have to do this without peeking at the previous slide to see the answer. What do you think will be our first move?

4. So we multiply both sides of the equation by (x + 2)(x – 1). Now we expand and compare the left side to the right side. If the left side and the right side are going to be equal then: A+B has to be 8 and -A+2B has to be 7.

5. A + B = 8 -A + 2B = 7 This gives us two equations in two unknowns. We can add the two equations and finish it off with back substitution. 3B = 15 B = 5 If B = 5 and A + B = 8 then A = 3. Cool!! But what does this mean?

6. Remember that our original mission was to break a big fraction into a couple of pieces. In particular to find A and B so that: We now know that A = 3 and B = 5 which means that And that is partial fraction decomposition!

10. Using the method of partial fractions to decompose

12. EXAMPLE 2: Find the partial fraction decomposition for 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.

13. Since the denominator is factorable we can pursue the decomposition. Because one of the factors in the denominator is quadratic, it is quite possible that its numerator could have an x term and a constant term—thus the use of Ax + B in the numerator. As in the first example, we multiply both sides of this equation by the LCD.

14. If the two sides of this equation are indeed equal, then the corresponding coefficients will have to agree: -1 = A + C 11 = 3A + B -10 = 3B + 4C On the next slide, we solve this system. We will start by combining the first two equations to eliminate A.

15. -1 = A + C 11 = 3A + B -10 = 3B + 4C 3 = -3A - 3C 11 = 3A + B Multiply both sides by -3 Add these two equations to eliminate A. 14 = B – 3C Multiply both sides of this equation by –3. Add this equation to eliminate B. -42 = -3B + 9C -10 = 3B + 4C -52 = 13C -4 = C We now have two equations in B and C. Compare the B coefficients. We can finish by back substitution. -1 = A + C-1 = A - 4A = 3 -10 = 3B + 4C -10 = 3B + 4(-4) -10 + 16 = 3B 2 = B

16. We have now discovered that A = 3, B = 2 and C = -4. OK, but I forgot what this means. Fair enough. We began with the idea that we could break the following fraction up into smaller pieces (partial fraction decomposition). Substitute for A, B and C and we are done.

20. EXAMPLE 3: For our next example, 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 A proceeds along the same lines as the previous example. Setup B considers that the second fraction could have come from two pieces.

21. Since we have already done an example with Setup A, this example will proceed with Setup B. Step 1 will be to multiply both sides by the LCD and simplify. Expand. Group like terms and factor. We now compare the coefficients of the two sides.

22. The last line of the previous slide left us here. If we compare the coefficients on each side, we have: A + B = 13 6A + 3B + C = 48 9A = 72 From the third equation A = 8. Substituting into the first equation: A + B = 13 so 8 + B = 13 and B = 5. Substituting back into the second equation: 6A + 3B + C = 48 so 6(8) + 3(5) + C = 48 48 + 15 + C = 48 63 + C = 48 and C = -15

23. To refresh your memory, we were looking for values of of A, B and C that would satisfy the partial fraction decomposition below and we did find that A= 8, B=5 and C=-15. So…..

25. EXAMPLE 4: 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.

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

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

28. From the previous slide we have that: If these two sides are equal then: 1 = A + B and 5 = 2A – 4B To eliminate A multiply both sides of the first equation by –2 and add. 2A – 4B = 5 -2A – 2B = -2 -6B = 3 so B = -1/2 If A + B = 1 and B = -1/2 then A –1/2 = 2/2 and A = 3/2

29. In summary then: