Partial Fractions

In an elementary algebra class, fractions with different denominators are combined into one fraction by getting a common denominator. An example is . The same technique holds for denominators with variables. An example of this is:

.

Now, what happens
if we want to reverse the process? That is, how would we decompose the above
rational fraction into separate fractions? Well, we would reverse the above steps.
But how do we determine that we should use 1 and 2 for numerators for the
individual fractions? Sure, we could guess numbers to see what works, but there
is a better, more systematic approach. The method is called **partial fraction decomposition**.

Think of a rational function as a gourmet meal. We are given the prepared dish (the rational function), but we want to know what ingredients (the separate fractions) were used to make the feast. Through some tests (partial fraction decomposition), we can determine the exact makeup of the meal.

There are many different ways to decompose a rational function into partial fractions. As a result, we shall examine four different cases. However, there are many more. But, from the following cases, a general pattern emerges to handle all situations.

**Case 1: Partial
Fractions with Distinct Linear Factors**

This technique
reverses the rational function that we examined above. Since we know the
answer, we can easily check our work to verify that we did the problem correctly.
These types of problem involve **linear
factors** since the denominator factors into terms of the form (*x* *a*)(*x* *b*)
and so on.

**Example 1**:

Decompose into partial fractions.

**Solution**:

We begin by factoring the denominator. The method of partial fractions is used when the denominator factors. After factoring, we write out something similar to our answer, but with variables in the numerator. It is our goal to solve for these unknowns.

.

Now, we want to
solve for *A* and *B*. To do this, we get a common denominator on both sides of the
equation.

.

Since the denominators are the same, we can just set the numerators equal to each other. Doing this, we get the following:

.

Our desire is to
solve for *A* and *B*. To do this, we pick values of *x* that will eliminate one of the variables. We begin by letting *x* = -1. (That will eliminate *A*.) Then we shall set *x* = -2. (That gets rid of *B*.)

Letting *x* = -1, we have .

Letting *x* = -2, we have .

Now, we plug in
those values for *A* and *B* into the first equation to get our
answer.

, as expected.

One might wonder what would happen if instead of putting , we had instead put . Would that change the answer?

The answer is no.
Instead of *A* = 1 and *B* = 2, we would have *A* = 2 and *B* = 1. The letters in the numerator are just variables. The order
does not matter. As a matter of convention, we typically start with *A* and continue through as many letters
are required.

**Case 2: Partial
Fractions with Some Repeated Linear Factors**

Now suppose one
of the factors is repeated *n* times. Students
often write out the term *n* times on
the bottom. Something like the following would appear on a student’s exam:

However, **this is not correct**!
If you were to following this technique, you would end up with the following
equations: *A* + *B* = 0 and *A* + *B*
= 6.

Clearly, both equations cannot be true. If they were then we would have 0 = 6. Something went wrong. The problem was in the factoring.

When using
partial fractions on repeated roots, write out the factor *n* times, but start the power at 1 and increase by 1 each time,
getting up to *n*. In the above
problem, one should write the following:

,
since *n* = 2.

Then, solve for the
variables *A* and *B* just as before in Example 1.

**Example 2**:

Using partial fractions, decompose .

**Solution**:

Notice that the linear factor (*x*) is repeated 3 times and the factor (*x* 1) is repeated 2 times. So, we set up the
following partial fractions:

.

Getting a common denominator, we have the following:

.

Now, we examine only the numerators.

Letting *x* = 1, we have:

Letting *x* = 0, we have:

But how are we
going to solve for *A*, *B*, and *D*. We cannot plug in numbers to cancel out the other variables. We
are going to have to solve a system of equations with three variables. To set
up the system, though, we need to pick three other values for *x*. Let use choose *x* = -1, *x* = 2, and *x* = -2. The work has been omitted, but
the results appear below.

Letting *x* = -1, we have:

Letting *x* = 2, we have:

Letting *x* = -2, we have:

Plugging in *C* = -1 and *E* = 4 into all three equations, we get the following equations:

(1)

(2)

(3)

Taking the negative of (2) and adding it to (1), we have:

(1*)

Taking -9 times (1) and adding it (3), we have:

(2*)

Adding (1*) and (2*), we have:

*B* = -2 means that *D* = -1. And using that information, we see that *A* = 1.

And so, after much labor, we see that:

.

**Case 3: Partial
Fractions with Distinct Irreducible Quadratic Factors**

So far, whenever
we have factored the denominator into pieces, we placed consecutive letters of
the alphabet above each fraction. This leads to a dangerous assumption: that
for any factoring, a letter is placed in the numerator. That is only true if
the factor in the denominator is a *linear
factor*. Something else happens with a **quadratic
factor**.

A quadratic
factor is anything of the form *ax*^{2}
+ *bx* + *c*, where *a* ≠ 0. Such a
factor is irreducible if the discriminate, *b*^{2}
4*ac*,
is less than 0.

In the case of
irreducible quadratic factors, instead of just putting a letter, we write a
polynomial that is one degree less than the denominator. In the case of linear
factors, one power less is just a constant. For a quadratic factor, one degree
less is a linear term, which is written as *Ax*
+ *B*. That goes in the numerator.

We are not going
to discussion factors larger than degree 2, but if there was a cubic in the
bottom, then we would write *Ax*^{2}
+ *Bx* + *C* in the numerator, and so on. Following this pattern, we could use
partial fractions on any type of factor that we might encounter.

**Example 3**:

Using Partial Fractions, show that .

**Solution**:

First we notice
that the denominator has already been factored. The first term, however, is a
quadratic factor, so we will have to write *Ax*
+ *B* in the numerator, not just *A*. The second factor, though, is linear,
so we can just write *C* in the
numerator. Doing this, we have the following:

.

Again, like before, we get a common denominator and then we only worry about setting the numerators equal to each other.

Letting *x* = 2, we have:

.

Letting *x* = 0 (and plugging in *C* = -1), we have:

Letting *x* = 1 (and plugging in *B* = -5 and *C* = -1), we have:

Plugging this information into the first equation, we have the following:

, which is precisely what we wanted to show.

**Case 4: Partial
Fractions with Some Repeated Irreducible Quadratic Factors**

This case is just like the Case 2, but with the linear factors in the numerator, not just constants, like we used in Case 3. Hopefully by this point, one sees the pattern that is emerging.

**Example 4**:

Set up, but to not solve, the partial fraction decomposition of .

**Solution**:

.

Following steps
similar to those done in Example 2, we could solve for all of the unknowns
(though we are not asked to do so). If we did, we would see that *A* = 2, *B* = 2, *C* = 3, *D* = -4, *E* = -3, *F* = 0, *G* = 3, *H* = -3, and *I* = -2.

If you look through all of the above examples, you will notice that in every example, the numerator has a smaller degree polynomial than the denominator. This was no accident.

Partial fractions can only be used on rational functions where the denominator has a larger degree than the numerator. In the case where the degree of the numerator is larger than degree of the denominator, we first use long division to create a fraction that we can use partial fractions on.

**Example 5**:

Use partial fractions to decompose .

**Solution**:

Notice that both the numerator and denominator have degree equal to 3. So, we have to use long division. First off, we have to expand the denominator. Doing that, we have:

.

Performing long division, we have the following:

And so, we get the following:

.

Now, we can use partial fractions on the new rational function. Since we have repeated linear factors, we follow the method used in Example 2.

.

Since the denominators are the same, we only look at the numerators.

.

Letting *x* = 1, we have:

Letting *x* = -3, we have:

Now,
unfortunately, we cannot plug in a value for *x* to solve directly for *A*.
However, since we know what *B* and *C* are, we can plug in any value for *x* and solve.

Letting *x* = 0, we have:

Plugging in *A*, *B*,
and *C*, we see:

And so, putting it all together, we have the following:

.