In this explainer, we will learn how to decompose rational expressions into partial fractions when the denominator has repeated linear factors.
For example, consider
When expanded, we see that has degree , while the degree of is less at 5. The result is that in this case, we can determine a partial fraction expansion like this:
Structure of Partial Fraction Expansion: Repeated Linear Factors
The partial fraction expansion is a sum of terms, where
- each term has the form ,
- is a constant,
- is one of the roots of , so corresponds to a distinct linear factor, and
- the exponent satisfies , where is the factor of corresponding to .
To see why this must be the case, consider with the degree of at most 2. Multiplying through by , for some constants , , and .
The first two terms are what you expect of the partial fraction expansion of in case the degree of is at most 1. Here, the degree may be 2, in which case long division gives a nonzero constant . Dividing both sides by gives and applying the partial fraction expansion to the middle term
If we set , on gathering the two like terms, we get
Extending this argument, we see that any rational expression of the form has the right kind of partial fraction expansion.
Here is a list of steps to follow to determine a partial fraction expansion, illustrated with this example: .
- If necessary, factorize the denominator (already done in this example). We note the form we expect: with unknown constants , , and . This is because the degree of the numerator here was less than 3. Otherwise, we would have begun with a long division.
- Multiplying this form with the denominator should give us the numerator . This is the equation
- We solve for these constants by substituting values of that allow us
to “pick off” values. For example, setting gives us
Setting gives us
For the middle term, we substitute any value we like, keeping in mind that we now know and . Setting , for example, and using the found values:
Hence, the partial fraction decomposition
Example 1: Decomposing Rational Expressions into Partial Fractions
Find and such that .
Here, the degree of is less than the degree of . Multiplying through by this gives the equation
We could set to its value and use an arbitrary value for and solve for in a second equation. But easier is to note that the equation above is of polynomials in . The leading coefficient of the right-hand side must be 2, so . Our solution is therefore
Suppose that we do not only have linear factors in the denominator. What can we expect?
The answer is that we treat irreducible factors of as if they were linear terms, except that instead of constants, we seek polynomials of lower degree than the factor, so for some constants , , , and that must be determined. The first two terms are as above for repeated linear factors. The last has a degree 1 numerator over the irreducible factor .
Here is a systematic way to solve this example. First, we write this as an equality of numerators, by multiplying the right-hand side by : and then substituting on both sides allows to solve for , since
Next, we rewrite this, substituting and maintaining the unexpanded terms, as an equation set to zero: where we have subtracted the value 3 at from both sides. This means that both sides are now equal to zero at , and so contain a factor , which we cancel:
The reason this helps is that now we can substitute once more to determine :
Finally, we can solve for and by setting in the original equation (with updated values of and ) to obtain just :
We can now substitute any other value for to determine . Say , which gives
We have found that