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Partial Fractions

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.

Well start by adding 2 fractions.

To find the partial fractions, we start with

The identity will be important for finding the

values of A and B.

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.

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.

This is where the identity is important.

What value would you substitute next ?

ANS Any value would do but x - 1 is good.

This is where the identity is important.

If we chose x 1 instead, we get 4 2A B,

giving the same result.

This is where the identity is important.

If we chose x 1 instead, we get 4 2A B,

giving the same result.

This is where the identity is important.

If we chose x 1 instead, we get 4 2A B,

giving the same result.

Its very important to write this each time

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.

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

The method weve used finds partial fractions for

expressions Ill call Type 1

where, the denominator has 2 linear factors,

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 )

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.

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

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

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

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.

Exercises

Express each of the following in partial

fractions.

1.

2.

3.

4.

Solutions

Check

Solutions

Check

Solutions

Check

( you dont need to write out the check in full )

Solutions

( I wont write out any more checks but it is

important to do them. )

Solutions

Solutions

If the denominator has 3 factors, we just extend

the method.

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.

Substituting B 3 gives A 1, an inconsistent

result

We need 3 constants if the degree of the

denominator is 3.

It would also be correct to write

but the fractions are not then reduced to the

simplest form

Using

Using

Using

There is no other obvious value of x to use so we

can choose any value.

Subst. for A and C

There is however, a neater way of finding B.

We had

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.

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 ).

So,

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

SUMMARY

- Check the answer by using a common denominator

for the right-hand side.

N.B. B can sometimes be zero.

Exercises

Solutions

Solutions

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

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.

e.g. 2

e.g. 2

e.g. 2

Exercise

1. Express the following in partial fractions

Solution

Dividing out

Partial Fractions

The 3rd type of partial fractions has a quadratic

factor in the denominator that will not factorise.

The partial fractions are of the form

Exercise

1. Express the following in partial fractions

Solution

Constants

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