Partial Fractions

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Partial Fractions
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In the presentation on algebraic fractions we saw
how to add 2 algebraic fractions.
To find partial fractions for an expression, we
need to reverse the process of adding fractions.
We will also develop a method for reducing a
fraction to 3 partial fractions.
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Well start by adding 2 fractions.
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To find the partial fractions, we start with
The identity will be important for finding the
values of A and B.
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To find the partial fractions, we start with
Multiply by the denominator of the l.h.s.
If we understand the cancelling, we can in future
go straight to this line from the 1st line.
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This is where the identity is important.
Why should I choose x 2 ?
ANS x 2 means the coefficient of B is zero, so
B disappears and we can solve for A.
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This is where the identity is important.
What value would you substitute next ?
ANS Any value would do but x - 1 is good.
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This is where the identity is important.
If we chose x 1 instead, we get 4 2A B,
giving the same result.
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This is where the identity is important.
If we chose x 1 instead, we get 4 2A B,
giving the same result.
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This is where the identity is important.
If we chose x 1 instead, we get 4 2A B,
giving the same result.
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Its very important to write this each time
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We never leave fractions piled up like this, so

• The halves are written in the denominators ( as
  2s ) and

• the minus sign is moved to the front of the 2nd
  fraction.

Finally, we need to check the answer.
A thorough check would be to reverse the process
and put the fractions together over a common
denominator.
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Another check is to use the cover-up method
To check A, find the value of x that makes the
factor under A equal to zero
( x 3 )
We get
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The method weve used finds partial fractions for
expressions Ill call Type 1
where, the denominator has 2 linear factors,
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The method weve used finds partial fractions for
expressions Ill call Type 1

• where,
• the denominator has 2 linear factors,

( we may have to factorise to find them )
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The method weve used finds partial fractions for
expressions Ill call Type 1

• where,
• the denominator has 2 linear factors,

• and the numerator is a polynomial of lower degree
  than the denominator

The degree of a polynomial is given by the
highest power of x.
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The method weve used finds partial fractions for
expressions Ill call Type 1

• where,
• the denominator has 2 linear factors,

• and the numerator is a polynomial of lower degree
  than the denominator

The degree of a polynomial is given by the
highest power of x.
1
Here the numerator is of degree
and the denominator of degree
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The method weve used finds partial fractions for
expressions Ill call Type 1
where, the denominator has 2 linear factors,
and the numerator is a polynomial of lower
degree then the denominator
The degree of a polynomial is given by the
highest power of x.
1
Here the numerator is of degree
2
and the denominator of degree
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The method weve used finds partial fractions for
expressions Ill call Type 1
where, the denominator has 2 linear factors,
and the numerator is a polynomial of lower
degree then the denominator
The degree of a polynomial is given by the
highest power of x.
1
Here the numerator is of degree
2
and the denominator of degree
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SUMMARY

• Multiply by the denominator of the l.h.s.

• Substitute a value of x that makes the
  coefficient of B equal to zero and solve for A.

• Substitute a value of x that makes the
  coefficient of A equal to zero and solve for B.

• Check the result by reversing the method or using
  the cover-up method.

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Exercises
Express each of the following in partial
fractions.
1.
2.
3.
4.
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Solutions
Check
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Solutions
Check
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Solutions
Check
( you dont need to write out the check in full )
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Solutions
( I wont write out any more checks but it is
important to do them. )
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Solutions
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Solutions
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If the denominator has 3 factors, we just extend
the method.
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The next type of fraction we will consider has a
repeated linear factor in the denominator.
This is wrong because the first 2 fractions just
give
, which is the same as having only one constant.
We will try this to see why it is also wrong.
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Substituting B 3 gives A 1, an inconsistent
result
We need 3 constants if the degree of the
denominator is 3.
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It would also be correct to write
but the fractions are not then reduced to the
simplest form
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Using
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Using
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Using
There is no other obvious value of x to use so we
can choose any value.
Subst. for A and C
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There is however, a neater way of finding B.
We had
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There is however, a neater way of finding B.
We had
Since this is an identity, the terms on each side
must be the same.
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There is however, a neater way of finding B.
We had
Since this is an identity, the terms on each side
must be the same.
We could also equate the coefficients of x ( but
these are harder to pick out ) or the constant
terms ( equivalent to putting x 0 ).
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So,
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So,
The cover-up method can only be used to check A
and C so for a proper check we need to put the
r.h.s. back over a common denominator.
We get
So the numerator gives
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SUMMARY

• Check the answer by using a common denominator
  for the right-hand side.

N.B. B can sometimes be zero.
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Exercises
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Solutions
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Solutions
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You may meet a question that combines algebraic
division and partial fractions.
In an exam you are likely to be given the form of
the partial fractions.
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Solution
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If you arent given the form of the partial
fractions, you just need to watch out for an
improper fraction.
You will need to divide out but you will probably
only need one stage of division so it will be
easy.
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e.g. 2
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e.g. 2
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e.g. 2
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Exercise
1. Express the following in partial fractions
Solution
Dividing out
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Partial Fractions
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The 3rd type of partial fractions has a quadratic
factor in the denominator that will not factorise.
The partial fractions are of the form
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Exercise
1. Express the following in partial fractions
Solution
Constants
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