Simplifying by factorising and cancelling
An algebraic fraction (also called a rational expression) is just a fraction whose numerator and denominator are algebraic expressions. To simplify one, you factorise the top and the bottom fully, then cancel any factor that appears in both.
- The golden rule of cancelling
- You may only cancel factors, never individual terms. A factor is something multiplied across the whole line; a term is something added. In \(\dfrac{(x-3)(x+3)}{(x+3)(x+4)}\) the bracket \((x+3)\) is a factor, so it cancels. You could never cancel the loose \(x\) values in \(\dfrac{x+3}{x+4}\).
Simplify \(\dfrac{x^2 - 9}{x^2 + 7x + 12}\).
Where this is assessed
This topic lives in Criterion A (applying factorising and fraction rules) and rewards Criterion C heavily: examiners want to see each factorising line and each cancelling step written clearly, not just a final fraction.
Multiplying and dividing
Multiplying rational expressions is the friendliest operation: factorise everything, cancel across the top and bottom of the whole product, then write what is left. There is no need for a common denominator.
Simplify \(\dfrac{x + 2}{x - 3} \times \dfrac{x^2 - 9}{x^2 + 4x + 4}\).
Dividing is one step longer: flip the second fraction and multiply. Never try to cancel until you have turned the division into a multiplication.
Simplify \(\dfrac{x^2 - 4}{x + 5} \div \dfrac{x - 2}{2x + 10}\).
Adding and subtracting
Adding and subtracting need a common denominator, exactly as with number fractions. The safe common denominator is the product of the two bottoms (or their lowest common multiple if they share a factor). Rewrite each fraction over that denominator, then combine the numerators. Take extra care with subtraction: the minus sign applies to every term of the second numerator.
Write \(\dfrac{2}{x} + \dfrac{3}{x + 1}\) as a single fraction.
The subtraction trap
For \(\dfrac{3}{x - 2} - \dfrac{1}{x + 1}\), the common denominator is \((x - 2)(x + 1)\), giving \(\dfrac{3(x + 1) - 1(x - 2)}{(x - 2)(x + 1)}\). The top becomes \(3x + 3 - x + 2 = 2x + 5\), because subtracting \(-2\) turns into \(+2\). Forgetting to distribute that minus over both terms is the single most common slip. The answer is \(\dfrac{2x + 5}{(x - 2)(x + 1)}\).
Factorise before you commit
Always factorise every denominator first. If they share a factor, the lowest common denominator is smaller than the product, and your numbers stay tidy. Then check the final numerator: if it factorises and cancels with the denominator, simplify once more.
Check yourself
Try each with the answers hidden. Show your factorising.
1. Simplify \(\dfrac{x^2 - 1}{x^2 - x}\). +
Factorise top and bottom: \(x^2 - 1 = (x - 1)(x + 1)\) and \(x^2 - x = x(x - 1)\). Cancel \((x - 1)\) to get \(\dfrac{x + 1}{x}\).
2. Simplify \(\dfrac{x}{x + 2} \times \dfrac{x^2 - 4}{x^2}\). +
Factorise \(x^2 - 4 = (x - 2)(x + 2)\), so the product is \(\dfrac{x\,(x - 2)(x + 2)}{(x + 2)\,x^2}\). Cancel \((x + 2)\) and one \(x\): \(\dfrac{x - 2}{x}\).
3. Write \(\dfrac{1}{x} + \dfrac{1}{x + 3}\) as a single fraction. +
Common denominator \(x(x + 3)\): \(\dfrac{(x + 3) + x}{x(x + 3)} = \dfrac{2x + 3}{x(x + 3)}\). The answer is \(\dfrac{2x + 3}{x(x + 3)}\).
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