Partial fractions

Posted in Maths, Core 4

Partial fractions involve splitting up a larger fraction into two or more parts. This technique is required to both make expressions easier to expand and make expressions easier to integrate.

There are 3 types of algebraic fractions that you need to be able to write as partial fractions.

Denominators in the form (ax + b)(cx + d)

Express Dynamic image 0 as partial fractions

First step is to write the equation as Dynamic image 1 where A and B are just some constant

Now multiply both sides by the denominator, in this case Dynamic image 2 to give Dynamic image 3

From here there are two methods to find the values of A and B, both methods have their merits and their uses and you'll find one is easier for you. In this example we'll use both to demonstrate how they work.

By substitution

The substitution method works by substituting a value for x into the equation to get rid of one of the constants. Basically put you want to make one of the brackets equal to 0.

Firstly to find the value for A we need to remove B which can be done by substituting x = 1 into the equation:

Dynamic image 4

Dynamic image 5

Dynamic image 6

Now to find B we need to remove A by substituting in x = -2:

Dynamic image 7

Dynamic image 8

Dynamic image 9

Now, returning to the original expression we wrote in step 1 we can substitute in values for A and B:

Dynamic image 10


By equating coefficients

This method is slightly more complex but is far more versatile. It involves equating the coefficients of x and the constant terms. Firstly expand the expression from step 2:

Dynamic image 11

Then equation the coefficients of x (the numbers and constants in front of the x) and the constants:

Dynamic image 12

Dynamic image 13

Now it's just a case of solving these simultaneous equations:

Dynamic image 14

Dynamic image 15

Dynamic image 16

So we get the same values from substitution.

Denominators in the form (ax + b)(cx2 + d)

Express Dynamic image 17 as partial fractions

The process is similar to above, firstly write the equation as Dynamic image 18 where A, B and C are just some constants

And again multiply both sides by the denominator to give: Dynamic image 19

By substituting x = -1 into the equation a value for A can be found:

Dynamic image 20

Dynamic image 21

Dynamic image 22

We can find B and C by equating the coefficients of x2, x and equating the constants. Firstly expand the right hand side of the equation:

Dynamic image 23

And equate the coefficients and constants:

Dynamic image 24

Dynamic image 25

Dynamic image 26

We know the value of A so we can find the value of B using the first equation:

Dynamic image 27

Now, substituting the value of B into the second equation we can find C:

Dynamic image 28

Finally we can check to see if we're right by substituting A and C into the last equation:

Dynamic image 29

Lastly we put our values of A, B and C into the original equation:

Dynamic image 30


Denominators in the form (ax + b)(cx + d)2

Express Dynamic image 31 as partial fractions

This time write the equation as Dynamic image 32 where A, B and C are just some constants

And again multiply both sides by the denominator to give: Dynamic image 33

By substituting x = 1 we can find a value for C:

Dynamic image 34

Dynamic image 35

By substituting x = -1 we can find a value for A:

Dynamic image 36

Dynamic image 37

By equating the coefficients of x2 and using the value for A above the value of B can be found:

Dynamic image 38

Putting these values in the original equation:

Dynamic image 39