Algebra · step 11 of 13

Rational expressions & partial fractions

Partial fractions run the addition of algebraic fractions backwards: given a single rational expression, split it into a sum with simpler denominators. The shape of the decomposition is dictated by the factors of the denominator — a linear factor gets a constant numerator, a repeated factor gets one term per power, and an irreducible quadratic gets a linear numerator.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Check the degree firstIf the numerator’s degree is at least the denominator’s, do polynomial division before decomposing.
Factor the denominator completelyThe decomposition is determined entirely by this factorisation.
Write the correct templateA/(x−a) for each distinct linear factor; A/(x−a) + B/(x−a)² for a squared factor; (Ax+B)/(x²+c) for an irreducible quadratic.
Solve for the constantsMultiply through by the denominator, then either substitute the roots (fast) or compare coefficients (systematic).

Worked example

Decompose (3x + 5)/((x − 1)(x + 2)) into partial fractions.

Template: A/(x−1) + B/(x+2).
Multiply through: 3x + 5 = A(x + 2) + B(x − 1).
Set x = 1: 8 = 3A, so A = 8/3.
Set x = −2: −1 = −3B, so B = 1/3.

Answer. (3x+5)/((x−1)(x+2)) = (8/3)/(x−1) + (1/3)/(x+2).

Where marks get dropped

These are the specific errors that cost credit on rational expressions & partial fractions questions — QED's rubric penalises each of them separately.

Decomposing an improper fraction directly. Divide first when the numerator’s degree is not smaller.
Using a constant numerator over an irreducible quadratic. It needs Ax + B, or the system will be inconsistent.
Omitting a term for a repeated factor. (x−1)² requires both A/(x−1) and B/(x−1)².

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Rational expressions & partial fractions — frequently asked questions

Why bother with partial fractions?

Because the pieces are easy to integrate, to sum (they often telescope), and to invert as Laplace transforms. It is a standard preparation step, not an end in itself.

Which method for the constants is faster?

Substituting the roots of the denominator kills most terms at once. Comparing coefficients is more reliable when repeated or quadratic factors are present.

What if the denominator does not factor?

Over ℝ every polynomial factors into linear and irreducible quadratic pieces, so a decomposition always exists — finding the factorisation may just be hard.

The rest of Algebra

Foundational algebra to close prerequisite gaps. Each subtopic below has its own method, worked example and mark-losing traps.

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