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.

    1 5 2 3 x x

    1 5 2 3 x x

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

2 7 8

2

   x x x                     2 2 1 5 2 3 1 1 x x x x x x

   

     2 1 10 5 3 3 x x x x 2 7 8

2

   x x x

2 7 8

2

   x x x

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:

1 2 2 7 8

2

       x B x A x x x

  

1 2   x x

Partial Fraction Decomposition Forms

1 2    x B x A   

1 2   x x

Repeated Linear Factors:

    2 7 8

2

x x x

Distinct Linear Factors:

 

2

1 1     x C x B x A   

1 1   x x x

     x x x x x

2 3 2

2 6 20 5

Partial Fraction Decomposition Forms

4

2 

  x C Bx x A

 

4

2 

x x

Repeated Quadratic Factors:

    x x x x 4 4 4 3

3 2

Distinct Linear and Quadratic Factors:

 

2 2 2

2 2      x D Cx x B Ax

  

2 2

2 2

  x x

    4 4 13 8

2 4 3

x x x x

2 7 8

2

   x x x

Find the partial fraction decomposition for:

  

                   1 2 1 2 7 8 x B x A x x x

  

1 2   x x

  

1 2   x x

     

1 2 1 1 2 2 7 8          x x x B x x x A x

  

1 2 1 2 7 8        x B x A x x x

   

2 1 7 8      x B x A x

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:

   

2 1 7 8      x B x A x B Bx A Ax x 2 7 8     

   

B A x B A x 2 7 8      

A+B has to be 8 and

• A+2B has to be 7.

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?

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:

1 2 2 7 8

2

       x B x A x x x

We now know that A = 3 and B = 5 which means that

1 5 2 3 2 7 8

2

       x x x x x

Now we will look at this same strategy applied to an LCD with one linear factor and one quadratic factor in the denominator. And that is partial fraction decomposition!

EXAMPLE 2: Find the partial fraction decomposition for

12 4 3 10 11

2 3 2

      x x 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.

  



12 4 3 12 4 3

2 3 2 3

       x x x x x x

   

 

3 4 3

2

    x x x

 



4 3

2 

  x x

Since the denominator is factorable we can pursue the decomposition.

       12 4 3 10 11

2 3 2

x x x x x

 



                3 4 3 4 10 11

2 2 2

x C x B Ax x x x x

 



3 4

2

  x x 3 4

2

    x C x B Ax

 



3 4

2

  x x

    10 11

2

x x

  

3   x B Ax

 

4

2 

 x C

C Cx B Bx Ax Ax 4 3 3

2 2

     

 

) 4 3 ( ) 3 ( 10 11

2 2

C B x B A x C A x x         

From this point…

  

 

4 3 10 11

2 2

        x C x B Ax x x

Use Judicious Substitution: Let x = -3:

     

 



 

 

 

4 3 3 3 3 10 3 11 3

2 2

             C B A

 

4 9 10 33 9      C C 13 52   4   C

Two reasons why using Judicious Substitution does not work for the other term:

• It has no real zeros.

0

• 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…

  

 

4 3 10 11

2 2

        x C x B Ax x x

Expand the right side completely: x2 terms:

 

) 4 3 ( ) 3 ( 10 11

2 2

C B x B A x C A x x         

C Cx B Bx Ax Ax 4 3 3

2 2

     

Collect like x-terms:

C B Bx Ax Cx Ax 4 3 3

2 2

     

     

C B x B A x C A 4 3 3

2

     

Associate like x-terms and factor out x: Now compare left- side terms with right-side terms: x terms: constants:

1   C A 11 3   B A 10 4 3    C B

1   C A 11 3   B A 10 4 3    C B

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 = -4: Substituting C into the first equation:

 

1 4     A 3  A

Substituting C into the third equation:

 

10 4 4 3     B 6 3  B 2  B

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

3  A 2  B 4   C

• 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 - 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 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).

 



3 4 4 3 10 11 12 4 3 10 11

2 2 2 2 3 2

                 x C x B Ax x x x x x x x x x

Substitute for A, B and C and we are done.

3 4 4 2 3

2

      x x x

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 2 2 3 2

3 72 48 13 9 6 72 48 13 9 6 72 48 13              x x x x x x x x x x x x x

There are two setups that we could use to begin: Setup A proceeds along the same lines as the previous example.

 

2 2 2

) 3 ( 3 72 48 13        x C Bx x A x x x x

Setup B considers that the second fraction could have come from two pieces.

 

2 2 2

) 3 ( 3 3 72 48 13         x C x B x A x x x x

 

 

   

 

           

 

   

A x C B A x B A x x Cx Bx Bx A Ax Ax x x Cx Bx Bx x x A x x Cx x Bx x A x x x x x C x x x B x x x A x x x x x C x B x A x x x x x x 9 3 6 72 48 13 3 9 6 72 48 13 3 9 6 72 48 13 3 3 72 48 13 3 3 3 3 3 72 48 13 3 ) 3 ( 3 3 72 48 13 3

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

                                                          

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.

   

A x C B A x B A x x 9 3 6 72 48 13

2 2

       

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

 

2 2 2

) 3 ( 3 3 72 48 13         x C x B x A x x x x

To refresh your memory, we were looking for values of

• f 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…..

 

2 2 2

) 3 ( 15 3 5 8 3 72 48 13          x x x x x x x

Our last example considers the possibility that the polynomial in the denominator has a smaller degree than the polynomial in the numerator.

EXAMPLE 4: Find the partial fraction decomposition for

8 2 5 15 4 2

2 2 3

     x x x x x

Since the order of the numerator is larger than the

• rder of the denominator, the first step is division.

5 16 4 2 8 2 5 2 5 15 4 2 8 2

2 3 2 2 3 2

            x x x x x x x x x x x x x

By long division we have discovered that:

8 2 5 2 8 2 5 15 4 2

2 2 2 3

          x x x x x x x x x

We will now do partial fraction decomposition on the remainder.

                        

B A x B A x B Bx A Ax x x B x A x x x x B x x x A x x x x B x A x x x x x x B x A x x x x x x 4 2 5 4 2 5 4 2 5 2 4 2 2 4 4 5 2 4 2 4 2 4 5 2 4 2 4 2 4 5 8 2 5

2

                                                   

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

From the previous slide we have that:

   

B A x B A x 4 2 5     

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

   

2 2 1 4 2 3 2 2 2 / 1 4 2 / 3 2 2 4 2 8 2 5 2 8 2 5 15 4 2

2 2 2 3

                          x x x x x x x B x A x x x x x x x x x x

In summary then:

You should now check out the companion piece to this tutorial, which contains practice problems, their answers and several complete solutions. Tips for partial fraction decomposition of N(x)/D(x): 1. If N(x) has a larger order than D(x), begin by long

• division. Then examine the remainder for

decomposition.

• 2. Factor D(x) into factors of (ax + b) and

 

c bx ax  

2

• 3. If the factor (ax + b) repeats then the decomposition

must include:

 

2

b ax B b ax A   

• 4. If the factor

decomposition must include:

c bx ax  

2

repeats then the

 

2 2 2

c bx ax D Cx c bx ax B Ax       