B: Algebra and functions: Decompose rational functions into partial fractions (denominators not more complicated than squared linear terms and with no more than 3 terms, numerators constant or linear).

The idea

Partial fractions reverse adding algebraic fractions: a rational function with a factorised denominator is rewritten as a sum of simpler fractions whose denominators are those factors. For distinct linear factors use one constant per factor, e.g. (2x+1)/((x−1)(x+2)) = A/(x−1) + B/(x+2). For a repeated linear factor include a term for each power, e.g. A/(x−1) + B/(x−1)^2. If the numerator is linear and the denominator has three linear factors, use three constants. Multiply through by the full denominator to form an identity valid for all x, then substitute convenient values (roots of factors) or compare coefficients. The method is assessed through correct form, accurate constants and valid algebraic manipulation.

What earns the marks

• Writes the correct partial fraction form, using one constant for each distinct linear factor and separate terms for repeated factors up to their power.
• Multiplies both sides by the full denominator to obtain an identity valid for all x, then solves for the unknown constants.
• Uses substitution of convenient x values, such as roots of the factors, to find constants efficiently; choosing a root makes all other terms vanish, which is the standard efficient method.
• Compares coefficients of powers of x when substitution alone is insufficient, for example with repeated factors.
• Handles numerators that are constant or linear correctly, including cases where the numerator has the same degree as a factor.
• Checks the decomposition by recombining the partial fractions or substituting a test value.

Common mistakes and how to put them right

• Using only one term for a repeated factor such as (x − 1)^2.

  Put it right by including both A/(x − 1) and B/(x − 1)^2.

• Forgetting to multiply every term by the full denominator when forming the identity.

  Put it right by writing the identity explicitly before substituting.

• Choosing an x value that is a root of a factor but then failing to substitute it into the full identity, or choosing a value that makes the algebra unnecessarily complicated.

  Put it right by substituting roots of the factors first, since this makes other terms vanish, and using coefficient comparison only for any remaining constants.

Exam tips

• Factorise the denominator fully before deciding the partial fraction form.
• Substitute roots of each factor first, then use coefficient comparison for any remaining constants.
• Show the identity clearly and substitute back one value to verify your constants.