Partial%20Fractions

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Partial Fractions
Splitting Fractions into Component fractions
Component-gt part of whole One into many
smaller!!
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In Calculus, there are several procedures that
are much easier if we can take a rather large
fraction and break it up into pieces. The
procedure that can decompose larger fractions is
called Partial Fraction Decompostition. We will
proceed as if we are working backwards through an
addition of fractions with LCD. EXAMPLE 1 For
our first example we will work an LCD problem
frontwards and backwards. Use an LCD to complete
the following addition.
The LCD is (x 2)(x 1). We now convert each
fraction to LCD status.
On the next slide we will work this problem
backwards
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Find the partial fraction decomposition for
As we saw in the previous slide the denominator
factors as (x 2)(x 1). We want to find
numbers A and B so that
The bad news is that we have to do this without
peeking at the previous slide to see the answer.
What do you think will be our first move?
Congratulations if you chose multiplying both
sides of the equation by the LCD. The good news
is that, since we are solving an equation, we can
get rid of fractions by multiplying both sides by
the LCD.
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So we multiply both sides of the equation by (x
2)(x 1).
Now we expand and compare the left side to the
right side.
If the left side and the right side are going to
be equal then AB has to be 8 and -A2B has to
be 7.
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This gives us two equations in two unknowns. We
can add the two equations and finish it off with
back substitution.
A B 8 -A 2B 7
3B 15 B 5
If B 5 and A B 8 then A 3.
Cool!! But what does this mean?
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Remember that our original mission was to break a
big fraction into a couple of pieces. In
particular to find A and B so that
We now know that A 3 and B 5 which means that
And that is partial fraction decomposition!
Now we will look at this same strategy applied to
an LCD with one linear factor and one quadratic
factor in the denominator.
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EXAMPLE 2 Find the partial fraction
decomposition for
First we will see if the denominator factors. (If
it doesnt we are doomed.)
The denominator has four terms so we will try to
factor by grouping.
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Since the denominator is factorable we can pursue
the decomposition.
Because one of the factors in the denominator is
quadratic, it is quite possible that its
numerator could have an x term and a constant
termthus the use of Ax B in the numerator.
As in the first example, we multiply both sides
of this equation by the LCD.
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If the two sides of this equation are indeed
equal, then the corresponding coefficients will
have to agree
-1 A C 11 3A B -10 3B 4C
On the next slide, we solve this system. We will
start by combining the first two equations to
eliminate A.
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Multiply both sides by -3
-1 A C 11 3A B -10 3B 4C
3 -3A - 3C 11 3A B
Add these two equations to eliminate A.
14 B 3C
Multiply both sides of this equation by 3.
Add this equation to eliminate B.
We now have two equations in B and C. Compare the
B coefficients.
We can finish by back substitution.
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We have now discovered that A 3, B 2 and C
-4.
Fair enough. We began with the idea that we could
break the following fraction up into smaller
pieces (partial fraction decomposition).
Substitute for A, B and C and we are done.
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EXAMPLE 3 For our next example, we are going to
consider what happens when one of the factors in
the denominator is raised to a power. Consider
the following for partial fraction decomposition
There are two setups that we could use to begin
Setup A proceeds along the same lines as the
previous example.
Setup B considers that the second fraction could
have come from two pieces.
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Since we have already done an example with Setup
A, this example will proceed with Setup B. Step 1
will be to multiply both sides by the LCD and
simplify.
Expand.
Group like terms and factor.
We now compare the coefficients of the two sides.
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The last line of the previous slide left us here.
If we compare the coefficients on each side, we
have A B 13 6A 3B C 48 9A 72
From the third equation A 8. Substituting into
the first equation A B 13 so 8 B 13 and
B 5.
Substituting back into the second equation 6A
3B C 48 so 6(8) 3(5) C 48 48 15 C
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63 C 48 and C -15
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To refresh your memory, we were looking for
values of of A, B and C that would satisfy the
partial fraction decomposition below and we did
find that A 8, B5 and C-15.
So..
Our last example considers the possibility that
the polynomial in the denominator has a smaller
degree than the polynomial in the numerator.
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EXAMPLE 4 Find the partial fraction
decomposition for
Since the order of the numerator is larger than
the order of the denominator, the first step is
division.
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By long division we have discovered that
We will now do partial fraction decomposition on
the remainder.
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Multiply both sides by
the LCD.
Distribute
Group like terms
Compare coefficients
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From the previous slide we have that
If these two sides are equal then 1 A B and
5 2A 4B
To eliminate A multiply both sides of the first
equation by 2 and add. 2A 4B 5 -2A 2B
-2 -6B 3 so B -1/2
If A B 1 and B -1/2 then A 1/2 2/2
and A 3/2
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In summary then
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3. If the factor (ax b) repeats then the
decomposition must include
You should now check out the companion piece to
this tutorial, which contains practice problems,
their answers and several complete solutions.