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Example 1 Determine
the partial fraction decomposition of each of the following.
(a)  [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
We’ll go through the first one in great detail to show the
complete partial fraction process and then we’ll leave most of the
explanation out of the remaining parts.
(a)
The first thing to do is factor the denominator as much as
we can.
So, by comparing to the table above it looks like the
partial fraction decomposition must look like,
Note that we’ve got different coefficients for each term
since there is no reason to think that they will be the same.
Now, we need to determine the values of A and B. The first step is to
actually add the two terms back up.
This is usually simpler than it might appear to be. Recall that we first need the least common
denominator, but we’ve already got that from the original rational
expression. In this case it is,
Now, just look at each term and compare the denominator to
the LCD. Multiply the numerator and
denominator by whatever is missing then add.
In this case this gives,
We need values of A
and B so that the numerator of the
expression on the left is the same as the numerator of the term on the
right. Or,
This needs to be true regardless of the x that we plug into this
equation. As noted above there are
several ways to do this. One way will
always work, but can be messy and will often require knowledge that we don’t
have yet. The other way will not
always work, but when it does it will greatly reduce the amount of work
required.
In this set of examples the second (and easier) method
will always work so we’ll be using that here.
Here we are going to make use of the fact that this equation must be true
regardless of the x that we plug
in.
So let’s pick an x,
plug it in and see what happens. For
no apparent reason let’s try plugging in . Doing this gives,
Can you see why we choose this number? By choosing we got the term involving A to drop out and we were left with a
simple equation that we can solve for B.
Now, we could also choose for exactly the same reason. Here is what happens if we use this value
of x.
So, by correctly picking x we were able to quickly and easily get the values of A and B. So, all that we need to
do at this point is plug them in to finish the problem. Here is the partial fraction decomposition
for this part.
Notice that we moved the minus sign on the second term
down to make the addition a subtraction.
We will always do that.
[Return to Problems]
(b)
Okay, in this case we won’t put quite as much detail into
the problem. We’ll first factor the
denominator and then get the form of the partial fraction decomposition.
In this case the LCD is and so adding the two terms back up give,
Next we need to set the two numerators equal.
Now all that we need to do is correctly pick values of x that will make one of the terms zero
and solve for the constants. Note that
in this case we will need to make one of them a fraction. This is fairly common so don’t get excited
about it. Here is this work.
The partial fraction decomposition for this expression is,
[Return to Problems]
(c)
In this case the denominator has already been factored for
us. Notice as well that we’ve now got
a linear factor to a power. So, recall
from our table that this means we will get 2 terms in the partial fraction
decomposition from this factor. Here
is the form of the partial fraction decomposition for this expression.
Now, remember that the LCD is just the denominator of the
original expression so in this case we’ve got . Adding the three terms back up gives us,
Remember that we just need to add in the factors that are missing
to each term.
Now set the numerators equal.
In this case we’ve got a slightly different situation from
the previous two parts. Let’s start by
picking a couple of values of x and
seeing what we get since there are two that should jump right out at us as
being particularly useful.
So, we can get A
and C in the same manner that we’ve
been using to this point. However,
there is no value of x that will
allow us to eliminate the first and third term leaving only the middle term
that we can use to solve for B. While this may appear to be a problem it
actually isn’t. At this point we know
two of the three constants. So all we
need to do is chose any other value of x
that would be easy to work with (
seems particularly useful here), plug that in along with the values of A and C and we’ll get a simple equation that we can solve for B.
Here is that work.
In this case we got this will happen on occasion, but do not
expect it to happen in all cases. Here
is the partial fraction decomposition for this part.
[Return to Problems]
(d)
Again, the denominator has already been factored for
us. In this case the form of the
partial fraction decomposition is,
Adding the two terms together gives,
Notice that in this case the second term already had the
LCD under it and so we didn’t need to add anything in that time.
Setting the numerators equal gives,
Now, again, we can get B
for free by picking .
To find A we
will do the same thing that we did in the previous part. We’ll use and the fact that we know what B is.
In this case, notice that the constant in the numerator of
the first isn’t zero as it was in the previous part. Here is the partial fraction decomposition
for this part.
[Return to Problems]
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