Algebraic expressions at A-Level
A-Level Maths assumes fluency with the algebraic manipulation you built at GCSE — index laws, surds and factorising — and builds on it immediately. This page extends those skills to full algebraic expressions and introduces algebraic fractions, which you'll use constantly in calculus, coordinate geometry and trigonometric identities throughout the course.
Four skills make up this topic:
- Index laws — applying the rules of indices to expressions with letters and numbers combined
- Expanding and factorising — including three brackets and factorising by grouping
- Algebraic fractions — simplifying, and adding, subtracting, multiplying and dividing them
- Surds — simplifying and rationalising denominators, including with a conjugate
Index laws for algebraic expressions
The rules of indices from GCSE apply exactly the same way to algebraic terms — you just handle the numerical coefficient and each letter's power separately.
| Rule | In symbols |
|---|---|
| Multiplying powers of the same base | \(a^m \times a^n = a^{m+n}\) |
| Dividing powers of the same base | \(a^m \div a^n = a^{m-n}\) |
| Power of a power | \((a^m)^n = a^{mn}\) |
| Negative index | \(a^{-n} = \dfrac{1}{a^n}\) |
| Fractional index | \(a^{1/n} = \sqrt[n]{a}\) |
Worked Example 1
Simplify \((2x^3y^2)^3 \div (4x^2y)^2\).
1
Expand each bracket using the power-of-a-power rule: \((2x^3y^2)^3 = 2^3x^9y^6 = 8x^9y^6\)
2
\((4x^2y)^2 = 4^2x^4y^2 = 16x^4y^2\)
3
Divide coefficients and subtract indices for each letter: \(8x^9y^6 \div 16x^4y^2 = \tfrac{8}{16}x^{9-4}y^{6-2}\)
4
Simplify: \(\tfrac{1}{2}x^5y^4\)
Answer\(\dfrac{x^5y^4}{2}\)
Expanding brackets
At GCSE you expanded pairs of brackets. At A-Level you need to expand three brackets fluently — expand two of them first, then multiply the result by the third.
Worked Example 2
Expand and simplify \((x + 1)(x + 2)(x + 3)\).
1
Expand the first two brackets: \((x + 1)(x + 2) = x^2 + 3x + 2\)
2
Multiply the result by the third bracket: \((x^2 + 3x + 2)(x + 3)\)
3
\(= x^3 + 3x^2 + 3x^2 + 9x + 2x + 6\)
4
Collect like terms: \(x^3 + 6x^2 + 11x + 6\)
Answer\(x^3 + 6x^2 + 11x + 6\)
Factorising expressions
Beyond the GCSE methods (common factor, quadratics, difference of two squares), A-Level introduces factorising by grouping for expressions with four terms: split into two pairs, factorise each pair, then factor out the common bracket.
Worked Example 3
Factorise fully: \(x^3 + 2x^2 + 3x + 6\).
1
Split into two pairs: \((x^3 + 2x^2) + (3x + 6)\)
2
Factorise each pair: \(x^2(x + 2) + 3(x + 2)\)
3
Both terms share the bracket \((x + 2)\) — factor it out
Answer\((x + 2)(x^2 + 3)\)
Simplifying algebraic fractions
To simplify an algebraic fraction, factorise the numerator and denominator fully, then cancel any bracket that appears in both. Never cancel individual terms that aren't common factors.
Worked Example 4
Simplify fully: \(\dfrac{x^2 - 9}{x^2 + x - 12}\).
1
Factorise the numerator (difference of two squares): \(x^2 - 9 = (x + 3)(x - 3)\)
2
Factorise the denominator: \(x^2 + x - 12 = (x + 4)(x - 3)\)
3
Cancel the common factor \((x - 3)\): \(\dfrac{(x + 3)(x - 3)}{(x + 4)(x - 3)}\)
Answer\(\dfrac{x + 3}{x + 4}\)
Adding, subtracting, multiplying and dividing algebraic fractions
Algebraic fractions follow the same rules as numerical fractions: for addition and subtraction, find a common denominator; for multiplication, multiply straight across (factorising and cancelling first where possible); for division, multiply by the reciprocal.
Worked Example 5
Express as a single fraction: \(\dfrac{2}{x + 1} + \dfrac{3}{x - 2}\).
1
Common denominator: \((x + 1)(x - 2)\)
2
Rewrite each fraction over it: \(\dfrac{2(x - 2) + 3(x + 1)}{(x + 1)(x - 2)}\)
3
Expand the numerator: \(2x - 4 + 3x + 3 = 5x - 1\)
Answer\(\dfrac{5x - 1}{(x + 1)(x - 2)}\)
Surds and rationalising denominators
You should already be confident simplifying surds and rationalising a denominator using a conjugate from GCSE. A-Level questions often phrase this as "show that ... can be written in the form \(a + b\sqrt{c}\)", asking you to state the integer values of \(a\) and \(b\) rather than just simplify.
Worked Example 6
Show that \(\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}\) can be written in the form \(a + b\sqrt{3}\), where \(a\) and \(b\) are integers. Find the values of \(a\) and \(b\).
1
Multiply top and bottom by the conjugate \((1 + \sqrt{3})\): \(\dfrac{(1 + \sqrt{3})(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}\)
2
Expand the numerator: \((1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3}\)
3
Expand the denominator: \((1 - \sqrt{3})(1 + \sqrt{3}) = 1 - 3 = -2\)
4
Divide every term by \(-2\): \(\dfrac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}\)
Answer\(a = -2, b = -1\)
Exam tips
💡 Tip 1
Factorise fully before you cancel
You can only cancel a bracket (a common factor), never an individual term inside one. Factorise the numerator and denominator completely first — cancelling too early is the single most common way this topic goes wrong.
💡 Tip 2
Multiply the whole numerator, not just one term
When combining fractions over a common denominator, every term in the original numerator must be multiplied by the same factor — a partial multiplication is a very common lost mark.
💡 Tip 3
Bracket a subtracted numerator
When subtracting one fraction from another, write the second numerator in brackets before expanding — e.g. \(3(x+2) - (2x - 2)\), not \(3(x+2) - 2x - 2\) — so the negative sign is distributed correctly across every term.
💡 Tip 4
Keep surd working exact
Never switch to a decimal approximation partway through a "show that" surd question — the mark scheme requires exact values throughout, ending in the exact \(a + b\sqrt{c}\) form asked for.
💡 Tip 5
Check by expanding back
After factorising, multiply your brackets back out — if you don't recover the original expression, you've made an error. This takes seconds and catches almost every sign mistake.
Common mistakes
Common Mistake 1
Cancelling terms instead of common factors
In \(\dfrac{x + 3}{x^2 + 3x}\), students sometimes try to cancel the "x" directly. Factorise the denominator first: \(x^2 + 3x = x(x + 3)\), then cancel the shared bracket \((x + 3)\): \(\dfrac{1}{x}\). Cancelling before fully factorising almost always gives a wrong answer.
Common Mistake 2
Forgetting the middle term when squaring a bracket
\((x + 3)^2\) is not \(x^2 + 9\) — the middle term is missing. Correctly expanded: \((x + 3)(x + 3) =\) \(x^2 + 6x + 9\). This mistake is especially costly in surd "show that" questions.
Common Mistake 3
Only distributing the negative sign partially
For \(\dfrac{3}{x - 1} - \dfrac{2}{x + 2}\), the numerator becomes \(3(x + 2) - 2(x - 1)\). Writing this as \(3x + 6 - 2x - 1\) forgets that the negative applies to both terms in \((x - 1)\). Correctly: \(3x + 6 - 2x + 2 =\) \(x + 8\).
Common Mistake 4
Rationalising only the denominator
When rationalising \(\dfrac{4}{2 + \sqrt{3}}\), students sometimes multiply only the denominator by the conjugate \((2 - \sqrt{3})\), forgetting the numerator must be multiplied too. Multiplying by \(\dfrac{2-\sqrt{3}}{2-\sqrt{3}}\) keeps the value unchanged — multiplying only the bottom changes it.
Common Mistake 5
Stopping before an expression is fully factorised
Factorising \(x^4 - 1\) as \((x^2 - 1)(x^2 + 1)\) is not complete — \((x^2 - 1)\) is itself a difference of two squares. Fully factorised: \((x - 1)(x + 1)(x^2 + 1)\). Always check whether a bracket can be factorised again.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Simplify \((3x^2y)^3\).Core Skills
Q2Simplify \(x^5y^2 \div x^2y\).Core Skills
Q3Expand and simplify \((x + 4)(x - 2)\).Core Skills
Q4Factorise fully: \(3x^2y + 6xy^2\).Core Skills
Q5Simplify \(\sqrt{72}\).Core Skills
Exam-Style
Q6Simplify \((2x^3y^4)^2 \div (4x^2y^3)\).Exam-Style
Q7Expand and simplify \((x + 1)(x + 2)(x - 3)\).Exam-Style
Q8Factorise fully: \(x^3 - 4x^2 + x - 4\).Exam-Style
Q9Simplify fully: \(\dfrac{x^2 - 16}{x^2 + 3x - 4}\).Exam-Style
Q10Rationalise the denominator of \(\dfrac{6}{\sqrt{5} - 2}\).Exam-Style
A* Challenge
Q11Simplify fully: \(\dfrac{x^2 - 1}{x + 3} \times \dfrac{x + 3}{x^2 + 2x + 1}\).A* Challenge
Q12Express as a single fraction in simplest form: \(\dfrac{3}{x - 1} - \dfrac{2}{x + 2}\).A* Challenge
Q13Show that \(\dfrac{2 + \sqrt{5}}{2 - \sqrt{5}}\) can be written in the form \(a + b\sqrt{5}\), stating the integer values of \(a\) and \(b\).A* Challenge
Q14Simplify fully: \(\dfrac{x^3 - 8}{x^2 - 4}\).A* Challenge
Q15Simplify fully, giving your answer as a single fraction: \(\dfrac{1}{x - 2} + \dfrac{1}{x + 2} - \dfrac{4}{x^2 - 4}\).A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(27x^6y^3\)
1
Cube each factor: \(3^3 = 27\), \((x^2)^3 = x^6\), \(y^3 = y^3\)
\(27x^6y^3\)
Q2 — \(x^3y\)
1
Subtract indices for each letter: \(x^{5-2} = x^3\), \(y^{2-1} = y\)
\(x^3y\)
Q3 — \(x^2 + 2x - 8\)
1
\((x + 4)(x - 2) = x^2 - 2x + 4x - 8\)
2
Collect like terms: \(x^2 + 2x - 8\)
\(x^2 + 2x - 8\)
Q4 — \(3xy(x + 2y)\)
1
HCF of \(3x^2y\) and \(6xy^2\) is \(3xy\)
2
\(3x^2y \div 3xy = x\); \(6xy^2 \div 3xy = 2y\)
\(3xy(x + 2y)\)
Q5 — \(6\sqrt{2}\)
1
Find the largest square factor of 72: \(72 = 36 \times 2\)
2
\(\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}\)
\(6\sqrt{2}\)
Exam-Style (Q6–Q10)
Q6 — \(x^4y^5\)
1
\((2x^3y^4)^2 = 4x^6y^8\)
2
\(4x^6y^8 \div 4x^2y^3 = x^{6-2}y^{8-3} = x^4y^5\)
\(x^4y^5\)
Q7 — \(x^3 - 7x - 6\)
1
\((x + 1)(x + 2) = x^2 + 3x + 2\)
2
\((x^2 + 3x + 2)(x - 3) = x^3 - 3x^2 + 3x^2 - 9x + 2x - 6\)
3
Collect like terms: \(x^3 + 0x^2 - 7x - 6\)
\(x^3 - 7x - 6\)
Q8 — \((x - 4)(x^2 + 1)\)
1
Split into pairs: \((x^3 - 4x^2) + (x - 4)\)
2
Factorise each pair: \(x^2(x - 4) + 1(x - 4)\)
3
Factor out the common bracket \((x - 4)\)
\((x - 4)(x^2 + 1)\)
Q9 — \(\dfrac{x - 4}{x - 1}\)
1
Factorise the numerator: \(x^2 - 16 = (x - 4)(x + 4)\)
2
Factorise the denominator: \(x^2 + 3x - 4 = (x + 4)(x - 1)\)
3
Cancel the common factor \((x + 4)\)
\(\dfrac{x - 4}{x - 1}\)
Q10 — \(6\sqrt{5} + 12\)
1
Multiply top and bottom by the conjugate \((\sqrt{5} + 2)\): \(\dfrac{6(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)}\)
2
Denominator: \((\sqrt{5})^2 - 2^2 = 5 - 4 = 1\)
3
\(\dfrac{6(\sqrt{5} + 2)}{1} = 6\sqrt{5} + 12\)
\(6\sqrt{5} + 12\)
A* Challenge (Q11–Q15)
Q11 — \(\dfrac{x - 1}{x + 1}\)
1
Factorise: \(x^2 - 1 = (x - 1)(x + 1)\); \(x^2 + 2x + 1 = (x + 1)^2\)
2
\(\dfrac{(x - 1)(x + 1)}{x + 3} \times \dfrac{x + 3}{(x + 1)^2}\)
3
Cancel \((x + 3)\) and one factor of \((x + 1)\)
\(\dfrac{x - 1}{x + 1}\)
Q12 — \(\dfrac{x + 8}{(x - 1)(x + 2)}\)
1
Common denominator \((x - 1)(x + 2)\): \(\dfrac{3(x + 2) - 2(x - 1)}{(x - 1)(x + 2)}\)
2
Expand: \(3x + 6 - 2x + 2 = x + 8\)
\(\dfrac{x + 8}{(x - 1)(x + 2)}\)
Q13 — \(a = -9, b = -4\)
1
Multiply by the conjugate \((2 + \sqrt{5})\): \(\dfrac{(2 + \sqrt{5})^2}{(2 - \sqrt{5})(2 + \sqrt{5})}\)
2
Numerator: \((2 + \sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}\)
3
Denominator: \(4 - 5 = -1\)
4
\(\dfrac{9 + 4\sqrt{5}}{-1} = -9 - 4\sqrt{5}\)
\(a = -9, b = -4\)
Q14 — \(\dfrac{x^2 + 2x + 4}{x + 2}\)
1
Factorise the numerator (difference of two cubes): \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)
2
Factorise the denominator: \(x^2 - 4 = (x - 2)(x + 2)\)
3
Cancel the common factor \((x - 2)\)
\(\dfrac{x^2 + 2x + 4}{x + 2}\)
Q15 — \(\dfrac{2}{x + 2}\)
1
Common denominator is \(x^2 - 4 = (x - 2)(x + 2)\): \(\dfrac{(x + 2) + (x - 2) - 4}{x^2 - 4}\)
2
Simplify the numerator: \(x + 2 + x - 2 - 4 = 2x - 4\)
3
Factorise and cancel: \(\dfrac{2(x - 2)}{(x - 2)(x + 2)} = \dfrac{2}{x + 2}\)
\(\dfrac{2}{x + 2}\)
Summary
- Index laws apply to algebraic terms exactly as they do to numbers — handle the coefficient and each letter's power separately.
- Expand three brackets by expanding two first, then multiplying by the third.
- Factorise expressions with four terms by grouping into pairs and factoring out a common bracket.
- To simplify an algebraic fraction, factorise fully first and cancel only common brackets — never individual terms.
- Add or subtract algebraic fractions over a common denominator; multiply straight across; divide by multiplying by the reciprocal.
- Rationalise a binomial denominator by multiplying top and bottom by its conjugate, keeping all working exact.
These skills are the toolkit for the rest of Pure Year 1 — you'll use index laws and algebraic fractions directly in Differentiation, and factorising throughout Quadratics and Equations & Inequalities as those pages go live.
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