Partial fractions split a single algebraic fraction into a sum of simpler fractions, reversing the process of adding them. There are three cases to know: distinct linear factors, a repeated factor, and improper fractions where you divide first. The constants are found by substituting convenient values of x or by comparing coefficients, and the results are used later for integration and binomial expansions.
Partial fraction forms
- Distinct linear factors: (px + q)/((x + a)(x + b)) = A/(x + a) + B/(x + b)
- Repeated factor: …/((x + a)(x + b)²) = A/(x + a) + B/(x + b) + C/(x + b)²
- Improper fraction: divide first, then split the proper part
Always check the degrees first: if the numerator's degree is not lower than the denominator's, divide before splitting.
What are partial fractions?
Adding 2/(x − 1) and 3/(x + 2) gives a single fraction. Partial fractions run that process backwards: given the single fraction, find the simpler ones that add to it. You need them to integrate rational functions and to expand some fractions as binomial series.
Distinct linear factors: the substitution method
Write one fraction for each linear factor, then multiply through by the full denominator so there are no fractions left. Substitute the value of x that makes each bracket zero: that removes every other term and leaves one equation in one constant.
For example, putting x = −a removes every term that contains (x + a), isolating the constant over that factor. It is fast and reliable for distinct linear factors.
Comparing coefficients
The second method is to expand the right-hand side after multiplying through, then equate the coefficients of each power of x and the constant terms. This gives simultaneous equations for the constants.
It is the dependable fallback when substitution is awkward, and a good way to check an answer you found by substitution.
Repeated factors
A squared factor such as (x − 2)² needs two fractions: one over (x − 2) and one over (x − 2)². Substituting the roots gives the constant over the other linear factor and the constant over the squared factor directly.
Substitution alone cannot find the constant over the single (x − 2), because x = 2 removes it. Find that one by comparing coefficients, usually of the highest power, or by substituting a convenient value such as x = 0.
Improper fractions
If the numerator's degree is equal to or greater than the denominator's, the fraction is improper. Divide first to get a quotient plus a proper fraction, then split the proper fraction. Skipping the division is the classic error: the standard forms only work for proper fractions.
Choosing the form and method
Read the denominator first to decide the form: distinct linear factors, a repeated factor, or an improper fraction. Then pick the quicker method for each constant: substitution where a root isolates it, comparing coefficients for the rest. Check your answer by substituting a value of x into both sides.
Worked examples
Example 1. Express (5x + 1)/((x − 1)(x + 2)) in partial fractions.
- Form: A/(x − 1) + B/(x + 2), so 5x + 1 = A(x + 2) + B(x − 1).
- x = 1: 6 = 3A, so A = 2.
- x = −2: −9 = −3B, so B = 3.
Answer: 2/(x − 1) + 3/(x + 2).
Marks: M1 for multiplying through, A1 for A and B.
Example 2. Express (4x + 6)/((x + 1)(x + 3)) in partial fractions by comparing coefficients.
- 4x + 6 = A(x + 3) + B(x + 1) = (A + B)x + (3A + B).
- x terms: A + B = 4. Constants: 3A + B = 6.
- Subtract: 2A = 2, so A = 1 and B = 3.
Answer: 1/(x + 1) + 3/(x + 3).
Marks: M1 for equating coefficients, A1 for A and B.
Example 3. Express (3x² − 3x + 3)/((x + 1)(x − 2)²) in partial fractions.
- Form: A/(x + 1) + B/(x − 2) + C/(x − 2)², so 3x² − 3x + 3 = A(x − 2)² + B(x + 1)(x − 2) + C(x + 1).
- x = −1: 9 = 9A, so A = 1.
- x = 2: 9 = 3C, so C = 3.
- x² coefficients: 3 = A + B, so B = 2.
Answer: 1/(x + 1) + 2/(x − 2) + 3/(x − 2)².
Marks: M1 for substitution, A1 for comparing to find B.
Example 4. Express (x² + 3x + 5)/((x − 1)(x + 2)) in the form D + A/(x − 1) + B/(x + 2).
- The denominator is x² + x − 2, so x² + 3x + 5 = (x² + x − 2) + (2x + 7). The quotient is D = 1.
- Split the remainder: 2x + 7 = A(x + 2) + B(x − 1).
- x = 1: 9 = 3A, so A = 3. x = −2: 3 = −3B, so B = −1.
Answer: 1 + 3/(x − 1) − 1/(x + 2).
Marks: M1 for dividing, A1 for the split.
Common mistakes
- Splitting an improper fraction without dividing. If the numerator's degree is not lower than the denominator's, divide first.
- Only one fraction for a repeated factor. (x − 2)² needs both B/(x − 2) and C/(x − 2)².
- Trying to find every constant by substitution. With a repeated factor, one constant must come from comparing coefficients or another value of x.
- Arithmetic slips when substituting. Substitute carefully and check the answer at a value such as x = 0.
- Not factorising the denominator fully. Factorise first, so you can see whether the factors are distinct or repeated.
How partial fractions are examined
In Edexcel International A-Level, partial fractions are Pure 4 content (paper WMA14), with a chapter of their own. In Edexcel A-Level (9MA0) they are second-year Pure. Questions ask you to:
- find A and B for distinct linear factors
- find A, B and C when one factor is repeated
- divide an improper fraction first, then split it
- use the partial fractions in a later part, to integrate or to expand as a series
Practise with the Pure 4 partial fractions exam questions and their mark schemes, and the partial fractions worksheet. The algebra revision engine has unlimited further questions.
Partial fractions are used in integration using partial fractions and in the binomial expansion for any power. Brush up first with algebraic fractions.
Frequently asked questions
When do you use partial fractions?
Mainly to integrate a rational function, and to find the binomial expansion of a fraction. Splitting it into simpler fractions makes each part easy to handle.
Which is better, substitution or comparing coefficients?
Substitution is quickest when the denominator has linear factors. Comparing coefficients always works and is needed for some constants with repeated factors. Many answers use both.
What do I do if the fraction is improper?
Divide first to get a polynomial plus a proper fraction, then split only the proper fraction into partial fractions.