2.8 Rational expressions and partial fractions — Learn

Prerequisites: factorising from 2.3 and 2.4; algebraic division and the factor theorem from 2.7; adding numerical fractions.

By the end of this section you can: simplify a rational expression by factorising and cancelling; add and subtract rational expressions; express a rational function as partial fractions with distinct or repeated linear factors; and state the values of $x$ that are excluded.

1. Rational expressions and their domain

A rational expression is a fraction of polynomials, such as

The expression is defined only where the denominator is not zero. The values that make the denominator zero are excluded values, and they stay excluded whatever simplification you make. This is the domain: the set of values the variable is allowed to take.

2. Simplifying by cancelling

To simplify, factorise the top and the bottom, then cancel a common factor. The cancellation is a division by the same expression, so it is valid wherever that factor is not zero.

Example 1. Simplify $\dfrac{x^2 - 1}{x^2 + 3x + 2}$.

Factorise both parts:

Cancel the common factor $x + 1$:

The simplified form is defined at $x = -1$, but the original expression is not: its denominator was zero there. Keep the excluded values $x = -1$ and $x = -2$ with the answer.

3. Adding and subtracting

A sum of rational expressions is put over a common denominator, exactly as with numbers.

Example 2. Write $\dfrac{2}{x} + \dfrac{3}{x + 1}$ as a single fraction.

The common denominator is the product $x(x + 1)$. Multiply the top and the bottom of each fraction to reach it:

Expand the numerator and collect terms:

4. Partial fractions

Partial fractions reverse the addition. A rational function is split into a sum of simpler fractions, each with one linear factor as its denominator. This is the form needed later for integration.

Example 3. Express $\dfrac{3x + 2}{(x - 1)(x + 2)}$ in partial fractions.

Write the unknown fractions with constant numerators:

Multiply both sides by the denominator:

Substitute the value of $x$ that makes each factor zero. At $x = 1$:

At $x = -2$:

So

Example 4. Express $\dfrac{2x + 3}{(x - 1)^2}$ in partial fractions.

A repeated factor needs a term for each power, up to the repeat:

Multiply through:

The two sides agree for every $x$, so the coefficients of like terms must match. The constant terms give $3 = -A + B$, and the $x$ terms give $2 = A$. So $A = 2$ and $B = 3 + A = 5$:

Example 5. Express $\dfrac{6x^2 + 5x - 5}{(x - 1)(x + 1)(x + 2)}$ in partial fractions.

Three distinct factors give three fractions:

Multiply through:

Substitute each root in turn. At $x = 1$, only $A$ survives:

At $x = -1$:

At $x = -2$:

So

Watch out. Two traps sit in partial fractions.

An improper fraction. If the degree of the numerator is not less than the degree of the denominator, divide first and split the remainder. For example, $\dfrac{x^2 + 1}{x - 1} = x + 1 + \dfrac{2}{x - 1}$.

A repeated factor given one term. For a denominator $(x - 1)^2$, one term $\dfrac{A}{x - 1}$ is not enough; you need $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$.

Re-check: add your partial fractions back over a common denominator and confirm you recover the original numerator. Check at one value of $x$ as well.

Progress check

1. Simplify $\dfrac{x^2 - 9}{x^2 - x - 6}$ and state the excluded values.
2. Write $\dfrac{1}{x - 1} - \dfrac{1}{x + 1}$ as a single fraction.
3. Express $\dfrac{3x + 2}{(x - 1)(x + 2)}$ in partial fractions.
4. Explain why a denominator of $(x + 2)^2$ needs two terms, not one.

You should now be able to

• state the excluded values of a rational expression;
• simplify by factorising and cancelling a common factor;
• add and subtract rational expressions over a common denominator;
• split a rational function into partial fractions with distinct linear factors;
• handle a repeated linear factor with one term for each power;
• divide first when the fraction is improper.