Partial Fractions (8.10)

How to decompose a rational expression into a rich compost of algebra and fractions

Start with a PODThe sum of this expression is

a. 1b. 0c. 1/xd. 2/(x-1)e. 2/(x2-1)

Everyone to the board!

1

1

1

1

xx

Start with a POD

So, answer e.

Today we’re learning that process in reverse and obtaining the partial fraction

decomposition.

1

x 1

1x 1

2

x2 1

Review

A rational expression is a polynomial divided by another polynomial

where f(x) and g(x) are polynomials.

f (x)

g(x)

New idea

f (x)

g(x)F1 F2 ... Fk

Where

In other words, rational expressions can be broken down (decomposed) into fractions of constants over linear expressions or linear expressions over quadratic expressions. (Remember that all polynomials can be factored into linear or quadratic factors with real coefficients.)

Fk A

px qor

Ax Bax2 bx c

New idea

f (x)

g(x)F1 F2 ... Fk

Where

Critical condition:f(x) must be a lower degree than g(x).We will use division to get there.

Fk A

px qor

Ax Bax2 bx c

The Method

1. Initial test: is f(x) a lower degree than g(x)? If not, use long division and decompose the remainder over the divisor.

2. Factor g(x)-- the denominator.

3. Split according to linear or quadratic factors.

4. Solve for the numerators-- A, B, C and so forth– by multiplying both sides by g(x).

Use it

4x2 13x 9x3 2x2 3x

Step 1. Is the numerator a lower degree than the denominator? In this case, yes.

Use it

Step 2. Factor the denominator.

4x2 13x 9x3 2x2 3x

4x2 13x 9x(x2 2x 3)

4x2 13x 9x(x 3)(x 1)

Use it

Step 3. Break into parts using the factors.

Use A, B, and C as the numerators.

4x2 13x 9x3 2x2 3x

4x2 13x 9x(x2 2x 3)

4x2 13x 9x(x 3)(x 1)

A

xB

x 3C

x 1

Use itStep 4. Solve for A, B, and C by multiplying each side by x3 + 2x2 - 3x.

Now what?

4x2 13x 9x3 2x2 3x

A

xB

x 3C

x 1

4x2 13x 9 A(x 3)(x 1) Bx(x 1)Cx(x 3)

Use it

Substitute values of x that make each factor

equal 0 in order to find A, B, and C.

If x = 0, then A = 3.

If x = 1, then C = 2.

If x = -3, then B = -1.

4x2 13x 9x3 2x2 3x

A

xB

x 3C

x 1

4x2 13x 9 A(x 3)(x 1) Bx(x 1)Cx(x 3)

Use it

If x = 0, then A = 3.

If x = 1, then C = 2.

If x = -3, then B = -1. And the final breakdown is

Test it.

4x2 13x 9x3 2x2 3x

A

xB

x 3C

x 1

4x2 13x 9 A(x 3)(x 1) Bx(x 1)Cx(x 3)

1

2

3

13

xxx

Use it a second time

This time you try the steps.

Notice how you have to do long division first.

4x3 x2 15x 292x3 x2 8x 4

2 x2 x 21

2x3 x2 8x 4

Use it a second timeWork with that remainder.

Notice the linear over quadratic fraction.

124

)12)(4(

21

)12(4)12(

21

)48()2(

21

482

21

2

2

2

2

2

23

2

23

2

x

C

x

BAx

xx

xx

xxx

xx

xxx

xx

xxx

xx

Use it a second timeSolve for A, B, and C.

If x = 1/2, then C = -5.If x = 0, then B = 1 (we can use the value of C we just got). Why did I

choose x=0?Then choose any other value for x to find A. I like x = 1.If x = 1, and we use the values of B and C, then A = 3.

x2 x 212x3 x2 8x 4

Ax Bx2 4

C

2x 1

x2 x 21(Ax B)(2x 1)C(x2 4)

Use it a second timeSolve for A, B, and C.

A = 3, B = 1, C = -5

The final breakdown:

x2 x 212x3 x2 8x 4

Ax Bx2 4

C

2x 1

x2 x 21(Ax B)(2x 1)C(x2 4)

2 3x 1x2 4

5

2x 1

Repeated linear factors

With a repeated linear factor, we have a special case.

Notice how we added a factor with the linear denominator to a second power.

22

2

)3()3()3(

3610

x

C

x

B

x

A

xx

xx

Repeated linear factors

In the case of a repeated linear factor, we have a special case. Here’s an example to illustrate.

A = -4, B = 5, C = 1

22

2

)3()3()3(

3610

x

C

x

B

x

A

xx

xx

Repeated quadratic factors

We have a similar situation with repeated quadratic factors.

A = 5, B = -3, C = 2, D = 0

5x3 3x2 7x 3(x2 1)2

Ax Bx2 1

Cx D(x2 1)2