# Mastering Partial Fractions: A step-by-step guide

The transition from lower secondary Mathematics to Additional Mathematics can be quite daunting for many students, with a distinct jump in the level of algebraic tenacity required. In this article, we will be going through the topic of partial fractions; how to break down an algebraic fraction step by step, as well as some things to keep in mind.

## Step 0: Before anything else, make sure you have a proper algebraic fraction

### What is a proper or improper fraction?

A proper fraction is a fraction where the numerator has a smaller numerical value than the denominator. The definition is slightly different when it comes to algebraic fractions. Instead of the numerical value, we compare the highest power of x. We only have a proper algebraic fraction if the highest power of x in the numerator is strictly less than in the denominator.

Here are a few examples:

$\frac{x-1}{x^2+3x+2}$ The highest power of x is 1 in the numerator and 2 in the denominator, thus it is a proper algebraic fraction.

$\frac{x^2+1}{2x^2+3x+1}$ The highest power of x is 2 in both the numerator and denominator, thus it is an improper algebraic fraction. Note that we are comparing the powers of x, not the coefficients.

### So what do we do when we have an improper algebraic fraction?

Convert to a mixed fraction of course. To do this, we will need to apply a long division of polynomials, which you'd have learnt in your polynomials class. Be careful of missing powers of x in this step! The image below shows where each part of the long division goes in conversion to mixed fractions.

$\begin{array}{r} Q(x)\phantom{)} \\ D(x){\overline{\smash{\big)}\,P(x) \phantom{)}}} \\. \\. \\. \\\overline{R(x)} \end{array} \rightarrow \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$

Now that we have a proper algebraic fraction to work with, we are ready to solve our partial fractions properly.

## Step 1: Factorise the denominator completely

This is the simplest step. At the O-level standard, we should be able to factorise any polynomial they give us. Do not forget the three algebraic identities that we learnt in Secondary 2.

## Step 2: Split the denominator up

For each factor you have in your denominator, we will write out a partial fraction according to which of the following three cases the factor falls under. On the right side of the equation, we will express the constants or coefficients in the numerator using capital letters, A, B and C.

### Case 1: Distinct linear factor

$\frac{px+q}{(ax+b)(cx+d)}=\frac{A}{ax+b}+\frac{B}{cx+d}$

The simplest case of the three, we just need to be able to recognise a linear factor ax+b.

Remember how we can only work with proper algebraic fractions? Each partial fraction

must also be a proper algebraic fraction, that’s why we put a constant A in the numerator.

### Case 2: Repeated linear factor

$\frac{px+q}{(ax+b)^2}=\frac{A}{ax+b}+\frac{B}{(ax+b)^2}$

This is the tricky one. Take note that $x^2$ falls under this case. Students usually lose marks when they forget to write both fractions when dealing with repeated linear factors. Lucky for us we will only be seeing the factor repeated once, though you can make a guess what would happen if the factor were cubed instead.

### Case 3: Quadratic factor that cannot be factorised

The $c^2$ in this case just means that we have $x^2$ plus a positive number. You can check that it really cannot be factorised by finding the discriminant. As for the numerator, since the highest power in the denominator is 2, we have $Bx+C$ to account for all possible proper algebraic fractions.

Putting this all together, here are a few examples of what it should look like. Note that (c) is an improper algebraic fraction.

(a) $\frac{x-9}{(x+1)(2x-3)}=\frac{A}{x+1}+\frac{B}{2x-3}$

(b) $\frac{2x^2+2x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$

(c) $\frac{x^3+2x}{(x-3)(x^2+2)}=1+\frac{3x^2+6}{(x-3)(x^2+2)}=1+\frac{A}{x-3}+\frac{Bx+C}{x^2+2}$

## Step 3: Solve for unknown constants

This involves multiple steps and different schools teach different methods. However, we find the smart substitution method the easiest to use. To show this, we will go through (b) from the examples above, describing the process along the way.

$\frac{2x^2+2x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$

First, multiply throughout by the denominator of the left-hand side. Take note of which factors cancel out on the right-hand side.

$2x^2+2x+1=Ax(x-1)+B(x-1)+Cx^2$

Next, substitute different values of $x$ such that each substitution leaves us with an equation with only one unknown for us to solve. Here, we can see that substituting $x = 1$ removes the terms containing A and B, and substituting $x = 0$ removes the terms containing A and C.

$x=1: 2+2+1=C(1)^2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C=5$

$x=0: 1=B(-1)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,B=-1$

Usually, when case 2 or 3 are involved, you will need to substitute x with any simple value in order to solve for the last unknown. Do not forget to substitute the values of unknowns that have already been found.

$x=2: 2(2)^2+2(2)+1=A(2)(1)-1(1)+5(2)^2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,13=2(A)+19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2A=-6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A=-3$

## Step 4: Conclusion Statement

We are almost done. Rewrite your equation from step 3, substituting the values of all unknowns.

$\frac{2x^2+2x+1}{x^2(x-1)}=-\frac{3}{x}-\frac{1}{x^2}-\frac{5}{x-1}$

Looking at the trend of questions set in exams, partial fractions are usually tested as a topic on its own, not mixed with the other topics of Additional Mathematics. The only topic it can be tested together with will be Integration in Secondary 4. However, these types of questions will usually be seen in school papers. The O-level paper will most likely test it as a question on its own, and it is usually worth 5 to 6 marks. Make sure to keep practising.

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