Partial Fractions by Richard Gill Supported in part by funding - PowerPoint PPT Presentation

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

EXAMPLE 1: For our first example we will work an LCD problem frontwards and backwards. Use an LCD to complete the following addition.  8 x 7 3 5       2 2 1 x x x x 2 The LCD is (x + 2)(x – 1). 3 5 We now convert each     fraction to LCD status. x 2 x 1       x 1 3 5 x 2               x 1 x 2 x 1 x 2     3 x 3 5 x 10 8 x 7         2 On the next slide we will work x 1 x 2 x x 2 this problem backwards

 Find the partial fraction decomposition for: 8 x 7   2 x x 2 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:  8 x 7 A B       2 x x 2 x 2 x 1      2 1 x x

Partial Fraction Decomposition Forms Distinct Linear Factors:  8 7 x A B       2 2 x x x 2 x 1      x 2 x 1 Repeated Linear Factors:   2 5 20 6 A B C x x          3 2 2 2 x x 1 x x x x 1      x x 1 x 1

Partial Fraction Decomposition Forms Distinct Linear and Quadratic Factors:    2 A Bx C 3 x 4 x 4  x  2   3 x 4 x 4 x   2  x x 4 Repeated Quadratic Factors:    3 Ax B Cx D 8 13 x x        2 2  4 2 x 2 2 4 4 x x x 2      2 2 x 2 x 2

 Find the partial fraction decomposition for: 8 x 7   2 x x 2  8 x 7 A B          2 1 2 1 x x x x            8 x 7 A B       x 2 x 1 x 2 x 1                x 2 x 1 x 2 x 1       A B        8 7 2 1 2 1 x x x x x   2 1 x x          8 x 7 A x 1 B x 2

Now we expand and compare the left side to the right side.          8 x 7 A x 1 B x 2      8 x 7 Ax A Bx 2 B           8 x 7 A B x A 2 B 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.

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

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:  8 x 7 A B       2 x x 2 x 2 x 1 We now know that A = 3 and B = 5 which means that  And that is partial 8 x 7 3 5   fraction     2 x x 2 x 2 x 1 decomposition! Now we will look at this same strategy applied to an LCD with one linear factor and one quadratic factor in the denominator.

EXAMPLE 2: Find the partial fraction decomposition for    2 11 10 x x    3 2 3 4 12 x x x 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.            3 2 3 2 x 3 x 4 x 12 x 3 x 4 x 12           2 x x 3 4 x 3    2     x 3 x 4

Since the denominator is factorable we can pursue the decomposition.     2 x 11 x 10 Ax B C        3 2 2 x 3 x 4 x 12 x 4 x 3       2 x 4 x 3       2    11 10 x x Ax B C           2  x 4 x 3     2 2   x 4 x 3 x 4 x 3   2            2 Ax B x 3 C x 4 x 11 x 10       2 2 Ax 3 Ax Bx 3 B Cx 4 C            2 2 x 11 x 10 A C x ( 3 A B ) x ( 3 B 4 C )

 0 From this point…              2 2 x 11 x 10 Ax B x 3 C x 4 Use Judicious Substitution: Let x = -3: 0                              2 2 3 11 3 10 3 3 3 3 4 A B C        9 33 10 C 9 4  52  13 C   C 4 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, A & B, that cannot be solved using a single equation.

Continuing on from this same point…              2 2 x 11 x 10 Ax B x 3 C x 4 Expand the right side completely:       2 2 Ax 3 Ax Bx 3 B Cx 4 C Collect like x-terms:       2 2 Ax Cx 3 Ax Bx 3 B 4 C Associate like x-terms and factor out x:             2 A C x 3 A B x 3 B 4 C Now compare left- side terms with right-side terms:            2 2 x 11 x 10 A C x ( 3 A B ) x ( 3 B 4 C )  C   x 2 terms: A 1  B  x terms: 3 A 11  C   constants: 3 B 4 10

Note how easily this system of equations can be solved using the single value we were able to obtain from Judicious Substitution?  C   A 1  B  3 A 11  C   3 B 4 10 Recall that C = -4: Substituting C into the first equation:       A 4 1  3 A Substituting C into the third equation:       3 B 4 4 10  3 B 6  B 2

Summarizing:  A 3  B 2   C 4 On the next slide, we solve this system of equations as if we did not know the value of C.

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

We have now discovered that A = 3, B = 2 and C = -4. OK, but I Fair enough. We began with the idea forgot what that we could break the following this means. fraction up into smaller pieces (partial fraction decomposition).    2 11 10 x x     3 2 3 4 12 x x x     2 x 11 x 10 Ax B C      Substitute for      2 2 3 4 4 3 x x x x A, B and C and we are done.   3 x 2 4     2 x 4 x 3

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:       2 2 2 13 x 48 x 72 13 x 48 x 72 13 x 48 x 72            2 3 2 2 x 6 x 9 x x 6 x 9 x x 3 There are two setups that we could use to begin: Setup A proceeds along    2 13 x 48 x 72 A Bx C   the same lines as the     2 2 previous example. ( 3 ) x x x x 3 Setup B considers that   2 13 48 72 x x A B C the second fraction         could have come from 2 2 x x 3 ( x 3 ) x x 3 two pieces.

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.         2   13 48 72   x x A B C        2 2 x x 3 x x 3        2 2   x x 3 ( x 3 ) x x 3 A   B   C           2 2 2 2 13 x 48 x 72 x x 3 x x 3 x x 3     2 x x 3 x 3     Expand.        2 2 13 x 48 x 72 A x 3 Bx x 3 Cx           2 2 2 Group 13 48 72 6 9 3 x x A x x Bx Bx Cx like         2 2 2 13 x 48 x 72 Ax 6 Ax 9 A Bx 3 Bx Cx terms             and 2 2 13 x 48 x 72 A B x 6 A 3 B C x 9 A factor. We now compare the coefficients of the two sides.