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

What this dot point is asking

SEAB wants you to take a single algebraic fraction whose denominator factorises into distinct linear factors and rewrite it as a sum of simpler fractions, one over each factor. This is the reverse of adding fractions over a common denominator. The skill matters because simpler fractions are far easier to integrate or expand, so partial fractions is a setup step for the calculus and binomial work.

The answer

What "proper" means

A fraction is proper when the degree (highest power) of the numerator is less than that of the denominator. The method here applies to proper fractions with a denominator that is a product of distinct linear factors. (If the fraction is improper, you would divide first, but that is beyond this dot point.)

Setting up the form

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

There is one unknown ($A$, $B$, and so on) per factor.

Clearing the denominators

Multiply both sides by the full denominator to get an identity with no fractions:

This is true for every value of $x$, which is what lets you choose convenient values next.

Finding the constants by substitution

Substitute the values of $x$ that make each factor zero, because each choice eliminates all but one unknown:

• Let $x = p$: the $B$ term vanishes, giving $A$ at once.
• Let $x = q$: the $A$ term vanishes, giving $B$.

This "cover-up" choice of values is the quickest way to find the constants. (Comparing coefficients also works but usually takes longer.)

Writing the answer

Put the constants back into the form. Always check by adding your partial fractions back over a common denominator; you should recover the original fraction.

Examples in context

Example 1. Preparing to integrate. A fraction like $\dfrac{1}{(x - 1)(x + 1)}$ is hard to integrate as it stands, but once split into $\dfrac{1}{2(x - 1)} - \dfrac{1}{2(x + 1)}$ each piece integrates to a logarithm. Partial fractions is the standard preparation for integrating algebraic fractions.

Example 2. Series and approximations. Breaking a fraction into simpler terms also makes it easier to expand each piece using the binomial theorem for an approximation. So partial fractions connects directly to the binomial work in the same module.

Try this

Q1. Write the partial-fraction form (with unknowns) for $\dfrac{1}{(x - 2)(x + 3)}$. [1 mark]

• Cue. $\dfrac{A}{x - 2} + \dfrac{B}{x + 3}$.

Q2. Express $\dfrac{4}{(x - 1)(x + 1)}$ in partial fractions. [3 marks]

• Cue. $4 = A(x + 1) + B(x - 1)$; $x = 1$ gives $A = 2$, $x = -1$ gives $B = -2$, so $\dfrac{2}{x - 1} - \dfrac{2}{x + 1}$.

Q3. Express $\dfrac{x}{(x + 1)(x + 2)}$ in partial fractions. [3 marks]

• Cue. $x = A(x + 2) + B(x + 1)$; $x = -1$ gives $A = -1$, $x = -2$ gives $B = 2$, so $\dfrac{-1}{x + 1} + \dfrac{2}{x + 2}$.