1.7iiPartial_Fractions

8/7/2019 1.7iiPartial_Fractions

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Partial Fractions

By

Mahatma Murthi



2

782

x x

xFind the partial fraction decomposition for:

122

782

!

x

B

x

A

x x

x



So we multiply both sides of the equation by (x + 2)(x ² 1).

Now we expand andcompare the left side tothe right side.

If the left side and the

right side are going tobe equal then:

A+B has to be 8 and

-A+2B has to be 7.

B A x B A x

B B x A A x x

x B x A x

x x x

B x x

x

A

x x

x x x

x x x

B

x

A

x x

x x x

278

278

2178

121

12212

7812

12122

781

2 2

!

!

!

!¹¹ º

¸©©ª

¨

¹ º

¸

©ª

¨

!



A + B = 8

-A + 2B = 7

This gives us two equations in two unknowns. We can addthe two equations and finish it off with back substitution.

3B = 15B = 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 bigfraction into a couple of pieces. In particular to find A

and B so that:

122

782

!

x

B

x

A

x x

x

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

1

5

2

3

2

78

2

!

x x x x

x

Now we will look at this same strategy applied to anLCD with one linear factor and one quadratic factor in

the denominator.

And that is partial

fractiondecomposition!



EXAMPLE 2: Find the partial fraction decomposition for

12431011

23

2

x x x

x x

First we will see if the denominator factors. (If itdoesn·t we are doomed.)

The denominator has four terms so we will try to factorby grouping.

43

3431243

2

223

!!

x x

x x x x x x



Since the denominator is factorable we can pursue thedecomposition.

3434

1011

1243

1011

22

2

23

2

!

!

x

C

x

B A x

x x

x x

x x x

x x

Because one of the factors in the denominator is quadratic,it is quite possible that its numerator could have an x termand a constant term³thus the use of Ax + B in thenumerator.

3434

1011

22

2

!

x x

B Ax

x x

x x As in the firstexample, wemultiply both sidesof this equation bythe LCD.



)43()3(1011

4331011

343

344

1011

343434

101134

22

222

22

2

2

2

22

22

C B x B A xC A x x

C C x B Bx Ax Ax x x

x x x

C x x

x

B Ax x x

x x x

C

x

B Ax

x x

x x x x

!

!

!

¹ º

¸©ª

¨

!

If the two sides of this equation are indeed equal, thenthe corresponding coefficients will have to agree:

-1 = A + C

11 = 3A + B

-10 = 3B + 4C

On the next slide, we solve thissystem. We will start bycombining the first twoequations to eliminate A.



-1 = A + C

11 = 3A + B-10 = 3B + 4C

3 = -3A - 3C

11 = 3A + B

Multiply both sides by -3

Add these two equationsto eliminate A.

14 = B ² 3C

Multiply bothsides of thisequation by ²3.

Add thisequation to

eliminate B.

-42 = -3B + 9C

-10 = 3B + 4C-52 = 13C

-4 = C

We now have two

equations in B and C.Compare the Bcoefficients.

We can finish byback 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 Iforgot whatthis means.

Fair enough. We began with the ideathat we could break the followingfraction up into smaller pieces(partial fraction decomposition).

34431011

1243

1011

22

2

23

2

!

!

xC

x B Ax

x x x x

x x x

x x

Substitute forA, B and C andwe are done.

3

4

4

23

2

!

x x

x



EXAMPLE 3: For our next example, we are going to considerwhat happens when one of the factors in the denominator is

raised to a power. Consider the following for partial fractiondecomposition:

22

2

2

23

2

3

724813

96

724813

96

724813

!

!

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 alongthe same lines as the

previous example. 22

2

)3(3

724813

!

x

C B x

x

A

x x

x x

Setup B considers thatthe second fractioncould have come from

two pieces.

22

2

)3(33

724813

!

x

C

x

B

x

A

x x

x x



A xC B A x B A x x

C x Bx Bx A Ax Ax x x

C x Bx Bx x x A x x

C x 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

936724813

396724813

396724813

33724813

33

33

3724813

3)3(33

7248133

22

222

222

22

2

2

222

2

22

22

!

!

!

!

!

¹¹ º

¸©©ª

¨

!

Since we have already done an example with Setup A, thisexample will proceed with Setup B. Step 1 will be to

multiply both sides by the LCD and simplify.

Expand.

Groupliketermsand

factor.We now compare the coefficients of the two sides.



A xC B A x B A x x 93672481322 !

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 thefirst 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



22

2

)3(33

724813

!

x

C

x

B

x

A

x x

x x

To refresh your memory, we were looking for values ofof A, B and C that would satisfy the partial fraction

decomposition below and we did find that A= 8, B=5 andC=-15.

So«..

22

2

)3(

15

3

58

3

724813

!

x x x x x

x x

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



EXAMPLE 4: Find the partial fraction decomposition for

82

515

42 2

23

x x

x x x

Since the order of the numerator is larger than the

order of the denominator, the first step is division.

5

1642

82

52

5154282

23

2

232

x

x x x

x x

x x

x x x x x



By long division we have discovered that:

82

5

282

515

42 22

23

!

x x

x

x x x

x x x

We will now do partial fraction decomposition on theremainder.



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

425

425

425

242

244

5

242424

524

2424

5

82

5

2

!

!

!

!

¹ º

¸©ª

¨

!

!

!

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 425 !

If these two sides are equal then:

1 = A + B and 5 = 2A ² 4B

To eliminate A multiply both sides of the firstequation by ²2 and add.

2A ² 4B = 5

-2A ² 2B = -2-6B = 3 so B = -1/2

If A + B = 1 and B = -1/2 then

A ²1/2 = 2/2 and A = 3/2



22

1

42

32

2

2/1

4

2/32

242

8252

8251542

22

23

!

!

!

!

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 thistutorial, which contains practice problems, theiranswers 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 fordecomposition.

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

3. If the factor (ax + b) repeats then the decompositionmust include:

2ba x

B

ba x

A

4. If the factor

decomposition must include:

cb xa x 2

repeats then the

222cb xax

DC x

cb xax

B Ax

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