# Partial Fractions – A Level Maths Revision

Beyond: Advanced walks you through the rhyme and reason behind partial fractions at A Level, complete with examples to cement your understanding.

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## Partial Fractions at A Level

Consider a problem where you are adding two algebraic fractions, for example: $\frac{3}{x\,+\,2}+\frac{5}{x\,-\,3}$

Like with any other fraction addition problem, you can find a common denominator (in this case, by cross multiplication – multiplying the numerator and denominator of each fraction by the denominator of the other fraction), then adding the numerators: $\frac{3}{x\,+\,2}+\frac{5}{x\,-\,3}$ $= \frac{3(x\,-\,3)}{(x\,+\,2)(x\,-\,3)}+\frac{5(x\,+\,2)}{(x\,+\,2)(x\,-\,3)}$ $= \frac{3x\,-\,9\,+\,5x\,+\,10}{(x\,+\,2)(x\,-\,3)}$ $= \frac{8x\,+\,1}{(x\,+\,2)(x\,-\,3)}$ $= \frac{8x\,+\,1}{x^2\,-\,x\,-\,6}$

Now, consider the opposite. You have a fraction, such as the result above, and want to break it apart or ‘decompose’ it into its constituent parts. There are numerous reasons you may want to do this, particularly when integrating or differentiating functions. The resulting decomposed fractions are called partial fractions.

Consider the following fraction: $\frac{3x\,-\,7}{(x\,+\,3)(x\,-\,5)}$

The denominator has two factors so you can expect to decompose this fraction into two partial fractions, with the denominators given by these two factors: $\frac{?}{x\,+\,3}+\frac{?}{x\,-\,5}$

How about the numerators? The numerator of the original fraction is 3x – 7. This must be the sum of the numerator of the first partial fraction, multiplied by x – 5, and the numerator of the second fraction, multiplied by x + 3.

The fact that the final numerator, 3x – 7, is a linear expression tells you something about the numerators of each partial fraction: they must both be constants. If either numerator had an algebraic term, the final numerator (which would be the result of multiplying two linear terms) would be quadratic. Call these constant numerators A and B: $\frac{A}{x\,+\,3}+\frac{B}{x\,-\,5}$

As the partial fractions should be true for all values of x, this relationship can be written as an identity: $\frac{3x\,-\,7}{(x\,+\,3)(x\,-\,5)} \equiv \frac{A}{x\,+\,3}+\frac{B}{x\,-\,5}$

Multiplying both sides of this identity by (x + 3)(x – 5) gives: $3x-7 \equiv A(x-5) + B(x+3)$

You can then expand this and compare the coefficients of the x terms and the constant terms: $3x - 7 \equiv Ax - 5A + Bx + 3B$ $3 = A + B$ (from the x terms) $-7 = -5A + 3B$ (from the constants)

These are two simultaneous equations in A and B – solving these gives: $A = 2$ $B = 1$

Substituting these back into the partial fractions gives: $\frac{3x\,-\,7}{x\,+\,3)(x\,-\,5)} \equiv \frac{2}{x\,+\,3}+\frac{1}{x\,-\,5}$

Example Question 1
Express the following fraction as partial fractions: $\boldsymbol{\frac{19x\,+\,11}{2x^2\,-x\,-\,6}}$

Start by factorising the denominator: $\frac{19x\,+\,11}{2x^2\,-x\,-\,6} = \frac{19x\,+\,11}{(2x\,+\,3)(x\,-\,2)}$

Given that the original numerator is linear and you are decomposing this fraction into just two partial fractions, both with linear denominators, the numerators of the partial fractions must be constants. $\frac{19x\,+\,11}{(2x\,+\,3)(x\,-\,2)} \equiv \frac{A}{2x\,+\,3} + \frac{B}{x\,-\,2}$

Multiply both sides by (2x + 3)(x – 2): $19x + 11 \equiv A(x - 2) + B(2x + 3)$ $19x + 11 \equiv Ax - 2A + 2Bx + 3B$

Comparing coefficients gives you: $A + 2B = 19$ $3B - 2A = 11$

Solve the simultaneous equations to get: $A = 5$ $B = 7$

Therefore: $\frac{19x\,+\,11}{2x^2\,-\,6} = \frac{5}{2x\,+\,3} + \frac{7}{x\,-\,2}$

Example Question 2
Express as partial fractions: $\boldsymbol{\frac{12x^2\,-\,30}{(x\,+\,2)(x\,-\,4)(2x\,+\,1)}}$

As the denominator is cubic, you will have three partial fractions: $\frac{12x^2\,-\,30}{(x\,+\,2)(x\,-\,4)(2x\,+\,1)} \equiv \frac{A}{x\,+\,2}+\frac{B}{x\,-\,4} + \frac{C}{2x\,+\,1}$

Multiply both sides by (x + 2)(x – 4)(2x + 1): $12x^2-30 \equiv A(x-4)(2x +1) + B(x+2)(2x+1) + C(x+2)(x-4)$

You could expand the brackets and compare coefficients to generate three simultaneous equations, hence finding the values of A, B and C. However, in this case it is simpler to substitute values of x to simplify the equation. By substituting three different values of x, you can remove each of the three sets of brackets.

Start by saying x = 4. This will remove the bracket (x – 4), as 4 – 4 = 0: $12(4)^2 - 30 = A(0)(9) + B(6)(9) + C(6)(0)$ $162 = 54B$ $B = 3$

Next, let x = -2: $12(-2)^2 - 30 = A(-6)(-3) + B(0)(-3) + C(0)(-6)$ $18 = 18A$ $A = 1$

Finally, let x = -0.5: $12(-0.5)^2 - 30 = A(-4.5)(0) + B(1.5)(0) + C(1.5)(-4.5)$ $-27 = -6.75C$ $C = 4$

Substituting back into the partial fraction gives: $\frac{12x^2\,-\,30}{(x\,+\,2)(x\,-\,4)(2x\,+\,1)} \equiv \frac{1}{x\,+\,2}+\frac{3}{x\,-\,4} + \frac{4}{2x\,+\,1}$

Example Question 3
Express as partial fractions: $\boldsymbol{\frac{4x^2\,+\,21x\,+\,30}{2x^2\,+\,9x\,+\,9}}$

The first thing to notice about this algebraic fraction is that it is improper. This is because the degree of the numerator (the highest power in the expression – x2) is the same as the degree of the denominator. To be a proper algebraic fraction, the degree of the numerator must be less than the degree of the denominator.

When faced with an improper fraction, start by converting it into a proper fraction. You can do this in the same way as you would any numeric improper fraction – divide the numerator by the denominator.

This tells us that $4x^2 + 21x + 30$ divided by $2x^2 + 9x + 9$ is 2, remainder $3x + 1$, or: $\frac{4x^2\,+\, 21x \,+ \,30}{2x^2\,+\,9x\, + \,9} = 2 + \frac{3x\, + \,12}{2x^2 \,+\,9x\, +\, 9}$

Now, consider just the proper fraction part. Start by factorising the denominator: $\frac{3x\, +\, 12}{2x^2\, +\, 9x\, +\, 9} = \frac{3x\, +\, 12}{(2x\, +\, 3)(x\, +\, 3)}$

Let: $\frac{3x\, +\, 12}{(2x\, +\, 3)(x\, +\, 3)} \equiv \frac{A}{2x\, +\, 3} + \frac{B}{x\, +\, 3}$

Multiply both sides by $(2x + 3)(x + 3)$ to get: $3x + 12 = A(x + 3) + B(2x + 3)$

Equating coefficients of x: $3 = A + 2B$

Equating constants: $12 = 3A + 3B$

Solving algebraically gives: $A = 5$ $B = -1$

So: $\frac{3x\, +\, 12}{(2x\, +\, 3)(x\, +\, 3)} \equiv \frac{5}{2x\, +\, 3} - \frac{1}{x\, +\, 3}$

Therefore: $\frac{4x^2\,+\,21x\,+\,30}{2x^2\,+\,9x\,+\,9} \equiv 2 + \frac{5}{2x\, +\, 3} - \frac{1}{x\, +\, 3}$

Example Question 4
Express as partial fractions: $\boldsymbol{\frac{9x^2\,+\,61x\,+\,108}{(x\,+\,3)^2(x\,+\,5)}}$

In this case, there is a repeated factor on the denominator. You have to take this into account when finding the partial fractions. The factors of $(x + 3)^2(x + 5)$ are $(x + 3)$, $(x + 3)^2$ and $(x + 5)$. This means you should resolve into three partial fractions, each with a constant numerator:

Let: $\frac{9x^2\,+\,61x\,+\,108}{(x\,+\,3)^2(x\,+\,5)} \equiv \frac{A}{x\,+\,3} + \frac{B}{(x\,+\,3)^2} + \frac{C}{x \,+ \,5}$

Multiply both sides by $(x + 3)^2(x + 5)$: $9x^2 + 61x + 108 = A(x+3)(x+5) + B(x + 5) + C(x + 3)^2$

Substitute x = -3: $9(-3)^2 + 61(-3) + 108 = A(0)(2) + B(2) + C(0)^2$ $6 = 2B$ $B = 3$

Substitute x = -5: $9(-5)^2 + 61(-5) + 108 = A(-2)(0) + B(0) + C(-2)^2$ $28 = 4C$ $C = 7$

Substituting in x-values allows you to find the values of B and C, but not A. If you consider the numerators, you can see that there are only two constants contributing towards the coefficient of x2: $9x^2 = Ax^2 + Cx^2$

Therefore: $9 = A + C$ $9 = A + 7$ $A = 2$

Giving the final partial fractions: $\frac{9x^2\,+\,61x\,+\,108}{(x\,+\,3)^2(x\,+\,5)} \equiv \frac{2}{x\,+\,3} + \frac{3}{(x\,+\,3)^2} + \frac{7}{x \,+ \,5}$

#### Practice Questions

Here are some to try yourself. Express each of the following as partial fractions.

1. $\frac{5x}{(x\,+\,2)(x\,-\,3)}$

2. $\frac{4x\,-\,28}{x^2\,-\,2x\,-\,15}$

3. $\frac{5x\,+\,7}{3x^2\,+\,10x\,+\,8}$

4. $\frac{220 \,-\,55x\,-\,11x^2}{(x\,-\,5)(x\,-\,3)(2x\,+\,5)}$

5. $\frac{6x^2\,+\,19x\,-\,2}{(x\,+\,4)(x\,+\,1)^2}$

1. $\frac{5x}{(x\,+\,2)(x\,-\,3)} \equiv \frac{2}{x\,+\,2}+\frac{3}{x\,-\,3}$
2. $\frac{4x\,-\,28}{x^2\,-\,2x\,-\,15} \equiv \frac{5}{x\,+\,3}-\frac{1}{x\,-\,5}$
3. $\frac{5x\,+\,7}{3x^2\,+\,10x\,+\,8} \equiv \frac{3}{2(x\,+\,2)}+\frac{1}{2(3x\,+\,4)}$
4. $\frac{220 \,-\,55x\,-\,11x^2}{(x\,-\,5)(x\,-\,3)(2x\,+\,5)} \equiv \frac{2}{x\,-\,3} + \frac{7}{2x\,+\,5}-\frac{11}{x\,-\,5}$
5. $\frac{6x^2\,+\,19x\,-\,2}{(x\,+\,4)(x\,+\,1)^2} \equiv \frac{2}{x\,+\,4}+\frac{4}{x\,+\,1}-\frac{5}{(x\,+\,1)^2}$