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1.3PureStretch

Partial Fractions

Partial fractions reverse the process of adding fractions: a single fraction with a factorised denominator is split back into a sum of simpler ones. This is essential for integration and binomial expansion later.

30 min Algebraic Methods
Edexcel A level Maths: 1.3 Partial FractionsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Split a fraction with distinct linear factors
  • Find the unknown constants (substitution method)
  • Use the cover-up method as a shortcut
  • Check the result by recombining
1

The form

A proper fraction whose denominator factorises into distinct linear factors\textbf{distinct linear factors} splits into a sum of fractions, one per factor, each with an unknown constant on top.

px+q(x+a)(x+b)=Ax+a+Bx+b\dfrac{px + q}{(x + a)(x + b)} = \dfrac{A}{x + a} + \dfrac{B}{x + b}
One term per linear factor, with constants AA, BB to find.
2

Finding the constants

Multiply both sides by the denominator to clear fractions, then substitute clever values of xx (each root) to isolate one constant at a time.

Example 1Express 5x+1(x+1)(x−2)\dfrac{5x + 1}{(x+1)(x-2)} in partial fractions.
1Write 5x+1=A(x−2)+B(x+1)5x + 1 = A(x-2) + B(x+1).
2Let x=2x = 2: 11=3B⇒B=11311 = 3B \Rightarrow B = \tfrac{11}{3}.
3Let x=−1x = -1: −4=−3A⇒A=43-4 = -3A \Rightarrow A = \tfrac{4}{3}.
Answer
4/3x+1+11/3x−2\dfrac{4/3}{x+1} + \dfrac{11/3}{x-2}

Tip — Substitute each root of the denominator — it kills all but one constant, solving it instantly.

3

The cover-up method

For distinct linear factors, you can find each constant directly: to get AA (over x+ax + a), cover up (x+a)(x + a) and evaluate the rest at x=−ax = -a. It is the substitution method done mentally.

Formula recap

px+q(x+a)(x+b)=Ax+a+Bx+b\dfrac{px+q}{(x+a)(x+b)} = \dfrac{A}{x+a} + \dfrac{B}{x+b}
Distinct linear factors.
substitute each root to find A,B\text{substitute each root to find } A, B
Solving the constants.
cover-up: evaluate at x=−a\text{cover-up: evaluate at } x = -a
Shortcut for each constant.

Common mistakes to avoid

Applying partial fractions to an improper fraction directly.
Divide first if the numerator degree ≥ denominator degree (see Algebraic Division).
Forgetting a term for each distinct factor.
There is exactly one partial fraction per linear factor.

Key takeaways

  • A proper fraction over distinct linear factors splits as A/(x+a) + B/(x+b).
  • Clear the denominator, then substitute each root to find the constants.
  • The cover-up method finds each constant quickly.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. px+q(x+a)(x+b)\dfrac{px+q}{(x+a)(x+b)} splits into:

    • Ax+a+Bx+b\dfrac{A}{x+a} + \dfrac{B}{x+b}
    • Ax+a×Bx+b\dfrac{A}{x+a} \times \dfrac{B}{x+b}
    • A(x+a)+B(x+b)A(x+a) + B(x+b)
    • A+B(x+a)(x+b)\dfrac{A+B}{(x+a)(x+b)}
    Show answer and explanation

    Answer: Ax+a+Bx+b\dfrac{A}{x+a} + \dfrac{B}{x+b}

    One term per factor.

  2. To find the constants, a quick approach is to substitute:

    • Each root of the denominator
    • x = 0 only
    • Large values
    • The numerator
    Show answer and explanation

    Answer: Each root of the denominator

    Each root isolates one constant.

  3. For 5x+1(x+1)(x−2)\dfrac{5x+1}{(x+1)(x-2)}, setting x=2x = 2 gives the equation for:

    • B
    • A
    • Both
    • Neither
    Show answer and explanation

    Answer: B

    x = 2 kills the A term.

  4. In that example, B=B =

    • 113\tfrac{11}{3}
    • 43\tfrac{4}{3}
    • 33
    • 1111
    Show answer and explanation

    Answer: 113\tfrac{11}{3}

    11 = 3B.

  5. Partial fractions require the fraction to be:

    • Proper (numerator degree < denominator)
    • Improper
    • A single factor
    • A square
    Show answer and explanation

    Answer: Proper (numerator degree < denominator)

    Otherwise divide first.

  6. The cover-up method finds A (over x + a) by evaluating the rest at:

    • x=−ax = -a
    • x=ax = a
    • x=0x = 0
    • x=1x = 1
    Show answer and explanation

    Answer: x=−ax = -a

    The root that makes x + a zero.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 5 lessons

Put the methods into practice: see how A-Level maths duels work. An account is required to play; the overview and sample are public.