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.

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 ax2 + bx + c, where a ≠ 0. Such a factor is irreducible if the discriminate, b2  4ac, 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 Ax2 + 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:

.