Partial Fractions | Best Introductory Guide
Introduction
When you did GCSE Maths, especially the HIgher paper and also year 12 A Level Maths, you will have done some work on single algebraic fractions.
An example of an algebraic fraction is: \frac{5}{(x+1)(x+2)}
What partial fractions allows us to do is to split a single algebraic fraction into two or more simpler fractions such as:
\frac{A}{(x+1)}+\frac{B}{(x+2)}Where A and B are both constants that can be determined. The reason for doing this is that it can help if there is work to be done with binomial expansion and it also makes performing integration much simpler.
Partial Fractions - Two Types
There are two types of partial fractions.
Type 1 is where there are two linear factors as mentioned above.
Type 2 is where there is a repeated linear factor.
We will look at both in turn.
Partial Fractions - Type 1
We will look at this by means of an example and discuss how the calculations have been done.
You can see that there are two linear factors in the denominator so we can express this algebraic fraction in the form:
\frac{A}{x+1}+\frac{B}{2 x-3}So what we have is essentially a mathematical identity where:
\frac{x-9}{(x+1)(2 x-3)} \equiv \frac{A}{x+1}+\frac{B}{2 x-3}The aim is determine the values of A and B and in order to do this we need to multiply throughout be the common denominator which is (x+1)(2 x-3) and this will give us:
x-9 \equiv A(2 x-3)+B(x+1)We need to pick suitable values of x that will make either A or B disappear.
\begin{aligned} x=-1:-10 & =-5 A \\ A & =2 \end{aligned}Once you have found one of the values you can then choose to let either x = 0 or x = 1 as this tends to be much quicker:
\begin{gathered} x=0:-9=2(-3)+B \\ -9=-6+B \\ B=-3 \end{gathered}Now the values of A and B have been found, we can now write our single algebraic fraction as follows:
\frac{x-9}{(x+1)(2 x-3)} \equiv \frac{2}{x+1}-\frac{3}{2 x-3}
Partial Fractions – Type 1
Let us consider another algebraic fraction question which we want to express as partial fractions.
You can see here that in the denominator there is a quadratic equation and this first of all needs to be factorised in order to give us two linear factors.
\frac{1}{x^2+x-2} \equiv \frac{1}{(x-1)(x+2)}We have two linear factors and this can now be split as follows:
\frac{1}{(x-1)(x+2)} \equiv \frac{A}{x-1}+\frac{B}{x+2}Again the aim is to determine the values of A and B, but first we need to multiply throughout by the common denominator and this will lead to:
1 \equiv A(x+2)+B(x-1)Now we need to find the values of A and B by picking values of x that will eliminate the values of A and B.
\begin{array}{ll} x=1: & 1=3 A \Rightarrow A=\frac{1}{3} \\ x=-2: & 1=-3 B \Rightarrow B=-\frac{1}{3} \end{array}We have two fractional values here, but take note how the answer has been written:
\frac{1}{(x-1)(x+2)} \equiv \frac{1}{3(x-1)}-\frac{1}{3(x+2)}Partial Fractions – Type 2
Consider the following example:
Again you will notice a quadratic in the denominator and this first of all needs to be factorised as follows:
\frac{3 x-1}{x^2-4 x+4} \equiv \frac{3 x-1}{(x-2)^2}Now what we have here is a repeated linear factor as shown:
\frac{3 x-1}{(x-2)^2} \equiv \frac{A}{x-2}+\frac{B}{(x-2)^2}Multiplying through by the common denominator gives:
3 x-1 \equiv A(x-2)+BAnd now the values of A and B can be determined:
\begin{array}{ll} x=2: & 5=B \\ x=0 & -1=-2 A+B \\ A=3 \end{array}And so we finally have:
\frac{3 x-1}{(x-2)^2} \equiv \frac{3}{x-2}+\frac{5}{(x-2)^2}You need to be always looking at the denominator when dealing with a type 1 or type 2 partial fraction question. The question is not going to tell you it is important that you are able to make that distinction and to then perform the necessary steps as needed.
Partial fractions is a very important topic and this is often extended into the areas of Binomial Expansion and also Integration. There are a number of past paper partial fractions questions that you can try and it is important that you are trying questions that test other subject areas such as the ones mentioned.
Try the examples once again on your own, if there are any uncertainties, but once you gain your confidence in this area, the topic is quite straightforward.