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 Algebra - Notes

## Partial Fractions

This section doesn’t really have a lot to do with the rest of this chapter, but since the subject needs to be covered and this was a fairly short chapter it seemed like as good a place as any to put it.

So, let’s start with the following.  Let’s suppose that we want to add the following two rational expressions.

What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add and/or subtract to get the original expression.  The process of doing this is called partial fractions and the result is often called the partial fraction decomposition.

The process can be a little long and on occasion messy, but it is actually fairly simple. We will start by trying to determine the partial fraction decomposition of,

where both P(x) and Q(x) are polynomials and the degree of P(x) is smaller than the degree of Q(x).   Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator.  That is important to remember.

So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible.  Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.

 Factor in denominator Term in partial fraction decomposition

Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let .  Also, it will always be possible to factor any polynomial down into a product of linear factors ( ) and quadratic factors ( ) some of which may be raised to a power.

There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples.  Speaking of which, let’s get started on some examples.

Now, we need to do a set of examples with quadratic factors.  Note however, that this is where the work often gets fairly messy and in fact we haven’t covered the material yet that will allow us to work many of these problems.  We can work some simple examples however, so let’s do that.

 Example 2  Determine the partial fraction decomposition of each of the following. (a)     [Solution] (b)     [Solution]     Solution (a)   In this case the x that sits in the front is a linear term since we can write it as,   and so the form of the partial fraction decomposition is,                                                   Now we’ll use the fact that the LCD is  and add the two terms together,                                          Next, set the numerators equal.                                               This is where the process changes from the previous set of examples.  We could choose  to get the value of A, but that’s the only constant that we could get using this method and so it just won’t work all that well here.   What we need to do here is multiply the right side out and then collect all the like terms as follows,                                              Now, we need to choose A, B, and C so that these two are equal.  That means that the coefficient of the x2 term on the right side will have to be 8 since that is the coefficient of the x2 term on the left side.  Likewise, the coefficient of the x term on the right side must be zero since there isn’t an x term on the left side.  Finally the constant term on the right side must be -12 since that is the constant on the left side.   We generally call this setting coefficients equal and we’ll write down the following equations.                                                                    Now, we haven’t talked about how to solve systems of equations yet, but this is one that we can do without that knowledge.  We can solve the third equation directly for A to get that .  We can then plug this into the first two equations to get,                                                   So, the partial fraction decomposition for this expression is,                                                  (b)   Here is the form of the partial fraction decomposition for this part.                                                   Adding the two terms up gives,                                            Now, set the numerators equal and we might as well go ahead and multiply the right side out and collect up like terms while we’re at it.                                        Setting coefficients equal gives,                                                                   In this case we got A and B for free and don’t get excited about the fact that .  This is not a problem and in fact when this happens the remaining work is often a little easier.  So, plugging the known values of A and B into the remaining two equations gives,                                            The partial fraction decomposition is then,
 Algebra - Notes