Topics

Rational expressions

Covers AS91261

What a rational expression is

A rational expression is an algebraic fraction — one polynomial divided by another:

$\frac{x^2 - 9}{x^2 + 5x + 6}$

They are named explicitly in the standard, and they behave exactly like numerical fractions. Every rule you know for $\frac{2}{3}$ applies.

Simplifying

Factorise the top and the bottom completely, then cancel common factors.
Worked through:

$\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x+3)(x-3)}{(x+3)(x+2)} = \frac{x-3}{x+2}$

You may only cancel whole factors, never individual terms.
- $\dfrac{x+3}{x+2}$ cannot be simplified — the $x$ 's are terms, not factors.
- $\dfrac{3x}{3y} = \dfrac{x}{y}$ can, because 3 is a factor of both.
The single most damaging error in this topic is cancelling across a plus sign. If you cannot see brackets around what you are cancelling, you cannot cancel it.

Multiplying

Factorise everything, cancel across the multiplication, then multiply what is left.

$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$

Worked through:

$\frac{x^2-4}{x+1} \times \frac{3x+3}{x-2} = \frac{(x+2)(x-2)}{(x+1)} \times \frac{3(x+1)}{(x-2)}$

$= 3(x+2)$

Cancel before multiplying out. Expanding first creates a mess that then has to be factorised again.

Dividing

Turn the second fraction upside down and multiply.

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$

Flip only the second fraction, and flip it before cancelling anything.

Adding and subtracting

These need a common denominator, exactly as with numbers.
- Factorise each denominator.
- Find the lowest common denominator.
- Rewrite each fraction with that denominator.
- Add or subtract the numerators, keeping the denominator.
- Simplify if the result factorises.
Worked through:

$\frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2x + 2 + 3x}{x(x+1)} = \frac{5x+2}{x(x+1)}$

Subtracting needs brackets. $\dfrac{A}{D} - \dfrac{B}{D} = \dfrac{A - (B)}{D}$ — the minus applies to every term of $B$ .

Restrictions on the variable

A fraction is undefined where its denominator is zero.
For $\dfrac{x-3}{x+2}$ , the expression is undefined at $x = -2$ .
Note that a cancelled factor still creates a restriction: $\dfrac{(x+3)(x-3)}{(x+3)(x+2)}$ is undefined at $x = -3$ and $x = -2$ , even though the simplified form only shows the second.

Solving equations with fractions

Multiply every term by the lowest common denominator to clear the fractions, then solve as usual.
Check your solutions. Any value that makes an original denominator zero must be rejected.
Worked through — solve $\dfrac{3}{x} + \dfrac{1}{2} = \dfrac{5}{x}$ :
- Multiply through by $2x$ : $6 + x = 10$
- So $x = 4$ , and since $4 \neq 0$ it is valid ✓

Tips and Tricks

Cancel factors, not terms. You may cancel $(x+3)$ from top and bottom because it is a factor. You may never cancel the $x$ from $\frac{x+3}{x+5}$ . Draw brackets around what you intend to cancel — if it does not sit inside brackets multiplying the rest, stop.

Factorise before you do anything else. Almost every rational expression question becomes easy once both parts are factorised, and almost none of them are possible before.

The 2025 report lists "factorised and cancelled simple polynomials" as a Merit behaviour, and "failed to manipulate expressions that involved surds, indices, or logs" among the Not Achieved. The technique is the mark.

Use brackets when subtracting fractions. $\frac{5}{x} - \frac{x+1}{x}$ is $\frac{5 - (x+1)}{x} = \frac{4-x}{x}$ , not $\frac{5-x+1}{x}$ .

Simplest algebraic form is required by the specification. If your answer still has a common factor top and bottom, it is not finished.

When you clear denominators to solve an equation, check each solution against the original denominators and reject any that makes one zero.

Worked ExampleSimplifying a rational expression

Simplify fully: $\dfrac{2x^2 + 5x - 3}{x^2 - 9} \div \dfrac{2x - 1}{x + 3}$ .

Step 1 — Change the division to a multiplication

Turn the second fraction upside down and multiply. Do this first, before any factorising, so you know what you are working with:

$\frac{2x^2 + 5x - 3}{x^2-9} \times \frac{x+3}{2x-1}$

Step 2 — Factorise the first numerator

$2x^2 + 5x - 3$ has a leading coefficient of 2, so use the splitting method:

$ac = 2 \times (-3) = -6$
Two numbers multiplying to $-6$ and adding to $5$ : 6 and $-1$
Split: $2x^2 + 6x - x - 3$
Pair: $2x(x+3) - 1(x+3)$
Factor: $(x+3)(2x-1)$

Step 3 — Factorise the first denominator

$x^2 - 9$ is a difference of two squares:

$x^2 - 9 = (x+3)(x-3)$

Step 4 — Rewrite everything in factored form

$\frac{(x+3)(2x-1)}{(x+3)(x-3)} \times \frac{(x+3)}{(2x-1)}$

Step 5 — Cancel common factors

Cancelling across the multiplication:

$(x+3)$ cancels once, top and bottom, from the first fraction
$(2x-1)$ cancels between the first numerator and the second denominator

$= \frac{(x+3)}{(x-3)}$

Step 6 — Check the result cannot simplify further

$\dfrac{x+3}{x-3}$ has no common factor. The $x$ terms cannot be cancelled — they are terms inside brackets, not factors of the whole expression.

Step 7 — State the restrictions

The expression is undefined where any original denominator is zero:

$x^2 - 9 = 0$ at $x = 3$ and $x = -3$
$2x - 1 = 0$ at $x = \frac{1}{2}$
$x + 3 = 0$ at $x = -3$ (from the divisor's numerator, which becomes a denominator)

So $x \neq 3$ , $x \neq -3$ , $x \neq \frac{1}{2}$ .