Partial fractions

Distinct linear factors

(5x + 1)/((x - 1)(x + 2)) ≡ A/(x - 1) + B/(x + 2). Multiply through: 5x + 1 ≡ A(x + 2) + B(x - 1). Three distinct factors need three fractions.

Finding the constants

• Substitute values of x that make a bracket zero (x = 1 gives 6 = 3A).
• Or compare coefficients of x², x and the constant.

The ≡ sign means the two sides are equal for every value of x, so both methods are allowed.

Repeated factors

A squared factor needs two fractions:

(x + 5)/((x - 1)(x + 2)²) ≡ A/(x - 1) + B/(x + 2) + C/(x + 2)².

Substitution finds A and C; compare the x² coefficients (or substitute another x) to find B.

Worked example

Express (5x + 1)/((x - 1)(x + 2)) in partial fractions.

1. 5x + 1 ≡ A(x + 2) + B(x - 1)
2. x = 1: 6 = 3A, so A = 2
3. x = -2: -9 = -3B, so B = 3

Answer: 2/(x - 1) + 3/(x + 2)

Check yourself

1. (3x + 7)/((x + 1)(x + 3)) ≡ A/(x + 1) + B/(x + 3). Find A.
   Show answer

   2. 3x + 7 ≡ A(x + 3) + B(x + 1). x = -1: 4 = 2A, so A = 2.

2. Write (5x - 4)/((x + 1)(x - 2)) as P/(x + 1) + Q/(x - 2). Find Q.
   Show answer

   2. 5x - 4 ≡ P(x - 2) + Q(x + 1). x = 2: 6 = 3Q, so Q = 2. (Also P = 3.)

3. What is the correct form of partial fractions for (x + 5)/((x - 1)(x + 2)²)?
   • A/(x - 1) + B/(x + 2) + C/(x + 2)²
   • A/(x - 1) + B/(x + 2)²
   • A/(x - 1)² + B/(x + 2)
   • A/(x - 1) + B/(x + 2)
   Show answer

   A/(x - 1) + B/(x + 2) + C/(x + 2)². A repeated factor needs both (x + 2) and (x + 2)² on the bottom.

4. (x + 5)/((x - 1)(x + 2)²) ≡ A/(x - 1) + B/(x + 2) + C/(x + 2)². Find B. Give your answer as a fraction in its simplest form.
   Show answer

   -2/3. x + 5 ≡ A(x + 2)² + B(x - 1)(x + 2) + C(x - 1). x = 1: 6 = 9A, so A = 2/3. Compare x² coefficients: 0 = A + B, so B = -2/3. (x = -2 gives C = -1.)

5. (6x² + 2x - 2)/(x(x - 1)(x + 2)) ≡ A/x + B/(x - 1) + C/(x + 2). Find C.
   Show answer

   3. 6x² + 2x - 2 ≡ A(x - 1)(x + 2) + Bx(x + 2) + Cx(x - 1). x = -2: 24 - 4 - 2 = 18 = 6C, so C = 3.

6. Why is it valid to substitute any value of x when finding partial fractions?
   • The equation only holds at the roots
   • x is always an integer
   • Partial fractions only work for x = 0
   • The two sides are identical, so they are equal for every value of x
   Show answer

   The two sides are identical, so they are equal for every value of x. The ≡ sign shows an identity, true for all x.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 4 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.