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CIE A-Level Maths Study Notes

2.1.4. Partial Fractions

Partial fractions are a technique used in algebra to break down complex rational expressions into simpler fractions that are easier to work with. This approach is particularly useful in integration, where simpler fractions can often be integrated more straightforwardly than a complex fraction.

Overview of Partial Fractions

Partial fractions involve expressing a given rational function as the sum of simpler rational functions. This technique is applicable when the degree of the numerator is less than the degree of the denominator.

Expressing Rational Functions in Partial Fractions

For a rational function ax+b(px+q)(rx+s)\frac{ax + b}{(px + q)(rx + s)}, it can be expressed in partial fractions as:

ax+b(px+q)(rx+s)Apx+q+Brx+s\frac{ax + b}{(px + q)(rx + s)} \equiv \frac{A}{px + q} + \frac{B}{rx + s}

To find the values of A and B:

1. Multiply through by (px+q)(px + q), substitute x=qpx = -\frac{q}{p}, and solve for A.

2. Multiply through by (rx+s)(rx + s), substitute x=srx = -\frac{s}{r}, and solve for B.

Decomposing Specific Denominators

For a more complex rational function such as ax2+bx+c(px+q)(rx+s)2\frac{ax^2 + bx + c}{(px + q)(rx + s)^2}, the decomposition would be:

ax2+bx+c(px+q)(rx+s)2Apx+q+Brx+s+C(rx+s)2\frac{ax^2 + bx + c}{(px + q)(rx + s)^2} \equiv \frac{A}{px + q} + \frac{B}{rx + s} + \frac{C}{(rx + s)^2}

To find A, B, and C:

1. Multiply through by (px+q)(px + q), substitute x=qpx = -\frac{q}{p}, and solve for A.

2. Multiply through by (rx+s)2(rx + s)^2, substitute x=srx = -\frac{s}{r}, and solve for C.

3. Substitute a constant (e.g., x=0x = 0) into the equation to find B.

Excluding Improper Fractions

When dealing with cases where the degree of the numerator is equivalent to or greater than that of the denominator, the fraction is termed 'improper'. In such cases, first convert the fraction into a proper fraction by dividing the denominator into the numerator and using the remainder in the partial fractions decomposition.


Express 4x27x1(x+1)(2x3)\frac{4x^2 - 7x - 1}{(x + 1)(2x - 3)} in partial fractions.


First, expand the brackets to simplify the expression: 4x27x12x2x3\frac{4x^2 - 7x - 1}{2x^2 - x - 3}

Since the highest power of x is the same in both the numerator and denominator, this is an improper fraction. Convert it into a proper fraction:

2+55x(x+1)(2x3) 2 + \frac{5 - 5x}{(x + 1)(2x - 3)}

Now, proceed to decompose the fraction:

Ax+1+B2x3=55x(x+1)(2x3)\frac{A}{x + 1} + \frac{B}{2x - 3} = \frac{5 - 5x}{(x + 1)(2x - 3)}A(2x3)+B(x+1)=55xA(2x - 3) + B(x + 1) = 5 - 5x

Solving for A when x=1x = -1:

5A=5+5    A=2-5A = 5 + 5 \implies A = -2

Solving for B when x=32x = \frac{3}{2}:

52B=5152    B=1\frac{5}{2}B = 5 - \frac{15}{2} \implies B = -1

Thus, the partial fraction decomposition is:

2+2x+1+12x32 + \frac{-2}{x + 1} + \frac{-1}{2x - 3}
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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