Express a proper rational fraction with distinct linear factors in the denominator as a sum of partial fractions

What this dot point is asking

SEAB wants you to reverse the process of adding fractions: to take a single proper rational fraction whose denominator factorises into distinct linear factors and split it back into a sum of simpler fractions, each with a constant numerator over one linear factor. This is essential preparation for integration and for series work at higher levels.

The answer

What proper means

A rational fraction is proper when the degree of the numerator is less than the degree of the denominator. Partial fractions in this form apply only to proper fractions; if the fraction is improper, you divide first (covered separately).

The form for distinct linear factors

For each distinct linear factor in the denominator, write one partial fraction with an unknown constant numerator:

There is one unknown per factor.

Clearing the denominators

Multiply both sides by the full denominator to get an identity free of fractions:

This must hold for all $x$, which lets you choose convenient values.

The cover-up substitution

Substitute the value of $x$ that makes one bracket zero; this eliminates the other unknown and gives one constant immediately. Repeat for the other root. This root-substitution method is the quickest route to the constants.

Checking the decomposition

Because a wrong constant is easy to produce, always verify the split before moving on. The quickest check is to recombine the partial fractions over the common denominator and confirm the numerator matches the original. A faster spot-check is to substitute one convenient value of $x$ (not a root) into both the original fraction and your decomposition and confirm they agree; if the original gives $\tfrac{1}{2}$ at $x = 0$ but your answer gives $\tfrac{1}{3}$, a constant is wrong. Building in a quick numerical check at a non-root value catches arithmetic slips that the substitution method can hide.

An alternative: comparing coefficients

Substituting the roots is fastest, but you can also find the constants by expanding the right side and equating coefficients of like powers of $x$. For $5x - 1 = A(x + 2) + B(x - 1)$, expanding gives $5x - 1 = (A + B)x + (2A - B)$, so $A + B = 5$ and $2A - B = -1$, a pair of simultaneous equations. This coefficient-comparison method is essential when a denominator includes a repeated or quadratic factor where root substitution alone cannot find every constant, so it is worth practising even on the simple linear case.

Examples in context

Example 1. Preparing to integrate. A fraction like $\dfrac{1}{(x - 1)(x + 2)}$ cannot be integrated as it stands, but once split into partial fractions each piece integrates to a logarithm, which is exactly why this skill leads into calculus.

Example 2. Decomposing a transfer relationship. In modelling how an output depends on an input through a product of simple factors, partial fractions separate the combined effect into independent contributions, each tied to one factor, making the behaviour easier to interpret.

Try this

Q1. Set up (do not solve) the partial fractions for $\dfrac{2x}{(x - 2)(x + 5)}$. [1 mark]

• Cue. $\dfrac{A}{x - 2} + \dfrac{B}{x + 5}$.

Q2. Express $\dfrac{1}{(x)(x + 1)}$ in partial fractions. [3 marks]

• Cue. $\dfrac{1}{x} - \dfrac{1}{x + 1}$.

Q3. Express $\dfrac{x + 4}{(x + 1)(x + 2)}$ in partial fractions. [4 marks]

• Cue. $x = -1$ gives $A = 3$; $x = -2$ gives $B = -2$: $\dfrac{3}{x + 1} - \dfrac{2}{x + 2}$.