Algebraic Fractions and Rationalising the Denominator
Simplifying, multiplying, dividing, adding and subtracting algebraic fractions, then rationalising the denominator using the conjugate. This is Extended-tier content, with worked examples and an exam-style question set out the way the mark scheme rewards.
0580 · E2.3 / E1.18 ExtendedSimplifying algebraic fractions
- Factorise the numerator fully.
- Factorise the denominator fully.
- Cancel any factor that appears in both.
Simplify \(\dfrac{x^2 + 3x}{x^2 - 9}\).
Factorise the top and bottom. The top is \(x(x + 3)\); the bottom is a difference of two squares, \((x + 3)(x - 3)\). Cancel the common factor \((x + 3)\):
\[\frac{x^2 + 3x}{x^2 - 9} = \frac{x(x + 3)}{(x + 3)(x - 3)} = \frac{x}{x - 3}\]You can cancel \((x + 3)\) because it multiplies the rest of the expression. You cannot cancel the \(x\) in \(\dfrac{x + 2}{x + 5}\), because there those \(x\) terms are added, not multiplied. Always factorise first, then cancel.
Multiplying and dividing
- Factorise every numerator and denominator.
- Cancel any common factors across the fractions.
- Multiply the numerators together and the denominators together.
- To divide: flip the second fraction first, then follow steps 1–3.
Simplify \(\dfrac{x + 2}{x} \times \dfrac{x^2}{x^2 - 4}\).
Factorise \(x^2 - 4 = (x + 2)(x - 2)\), then cancel \((x + 2)\) and one factor of \(x\):
\[\frac{x + 2}{x} \times \frac{x^2}{(x + 2)(x - 2)} = \frac{x}{x - 2}\]Simplify \(\dfrac{3}{x + 1} \div \dfrac{6}{x^2 - 1}\).
Multiply by the reciprocal, and factorise \(x^2 - 1 = (x + 1)(x - 1)\):
\[\frac{3}{x + 1} \times \frac{(x + 1)(x - 1)}{6} = \frac{x - 1}{2}\]Adding and subtracting
- Find the common denominator (usually the product of the two denominators).
- Rewrite each fraction over that denominator.
- Combine the numerators, keeping signs carefully.
- Expand and simplify the numerator.
Write \(\dfrac{2}{x} + \dfrac{3}{x + 1}\) as a single fraction.
The common denominator is \(x(x + 1)\):
\[\frac{2(x + 1)}{x(x + 1)} + \frac{3x}{x(x + 1)} = \frac{2x + 2 + 3x}{x(x + 1)} = \frac{5x + 2}{x(x + 1)}\]Write \(\dfrac{5}{x - 2} - \dfrac{3}{x + 2}\) as a single fraction.
The common denominator is \((x - 2)(x + 2)\). Keep the subtraction sign attached to the whole second numerator:
\[\frac{5(x + 2) - 3(x - 2)}{(x - 2)(x + 2)} = \frac{5x + 10 - 3x + 6}{(x - 2)(x + 2)} = \frac{2x + 16}{(x - 2)(x + 2)}\]When subtracting, the minus sign applies to every term in the second numerator. Using a bracket keeps it right: \(-3(x - 2) = -3x + 6\), not \(-3x - 6\).
Surds: a quick recap
A surd is a root that does not simplify to a whole number, such as \(\sqrt{2}\) or \(\sqrt{5}\). The two rules you need for rationalising are:
\[\sqrt{a} \times \sqrt{b} = \sqrt{ab} \qquad \sqrt{a} \times \sqrt{a} = a\]Simplify a surd by taking out the largest square factor.
Simplify \(\sqrt{50}\).
Rationalising a single-term denominator
Rationalising means rewriting a fraction so the denominator has no surd. When the denominator is a single surd \(\sqrt{a}\), multiply the top and bottom by \(\sqrt{a}\), because \(\sqrt{a} \times \sqrt{a} = a\).
Rationalise \(\dfrac{6}{\sqrt{3}}\).
Multiply the top and bottom by \(\sqrt{3}\):
\[\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\]Rationalising a two-term denominator
- Write down the conjugate: same two terms, opposite sign between them.
- Multiply both numerator and denominator by the conjugate.
- Expand the denominator using \((a+b)(a-b)=a^2-b^2\) to clear the surd.
- Expand the numerator and simplify.
Rationalise \(\dfrac{1}{1 + \sqrt{3}}\).
Multiply the top and bottom by the conjugate \(1 - \sqrt{3}\):
\[\frac{1}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{1 - \sqrt{3}}{1^2 - (\sqrt{3})^2} = \frac{1 - \sqrt{3}}{1 - 3}\]Simplify the denominator and tidy the signs:
\[= \frac{1 - \sqrt{3}}{-2} = \frac{\sqrt{3} - 1}{2}\]Exam-style question
(a) Simplify \(\dfrac{x^2 - 16}{x^2 + 5x + 4}\). [3]
(b) Write \(\dfrac{4}{x + 2} + \dfrac{1}{x - 3}\) as a single fraction. [2]
(c) Rationalise the denominator of \(\dfrac{10}{\sqrt{5}}\). [2]
Show solution (a)
Factorise both: \(x^2 - 16 = (x + 4)(x - 4)\) and \(x^2 + 5x + 4 = (x + 4)(x + 1)\). Cancel \((x + 4)\):
\[\frac{x^2 - 16}{x^2 + 5x + 4} = \frac{(x + 4)(x - 4)}{(x + 4)(x + 1)} = \frac{x - 4}{x + 1}\]Show solution (b)
Common denominator \((x + 2)(x - 3)\):
\[\frac{4(x - 3) + 1(x + 2)}{(x + 2)(x - 3)} = \frac{4x - 12 + x + 2}{(x + 2)(x - 3)} = \frac{5x - 10}{(x + 2)(x - 3)}\]Show solution (c)
Multiply the top and bottom by \(\sqrt{5}\):
\[\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}\]Algebraic fractions and surds are core Extended skills that feed into solving equations, functions and the non-calculator paper. Always factorise fully before cancelling, keep the subtraction sign attached to the whole numerator, and remember that a two-term denominator is rationalised with its conjugate and the difference of two squares.
Common mistakes
You can only cancel a factor that multiplies the whole numerator and denominator. In \(\dfrac{x + 2}{x + 5}\) nothing cancels, because the \(x\) values are terms, not factors.
To simplify a fraction you must factorise the top and bottom fully before looking for common factors. Cancelling before factorising leads to wrong answers.
When subtracting fractions, the minus sign applies to every term of the second numerator. Use a bracket: \(- (3x - 6) = -3x + 6\).
A final answer should have a rational denominator and any surds fully simplified, so \(\sqrt{50}\) becomes \(5\sqrt{2}\) and no root is left underneath.
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