GCSE Higher — Algebra
This is the algebra practice for GCSE Higher, the content area that carries 30% of the paper. Coverage is complete: all 25 algebra statements, each with practice that matches how the topic is actually examined. Among them: algebraic notation, substitution into formulae, expressions, equations, formulae, identities and inequalities, simplifying, expanding and factorising, rearranging formulae and identities, equivalence and algebraic proof. Higher only: translations and reflections of graphs, gradients and areas under graphs, equation of a circle and its tangent and iteration. A Foundation entry is never asked for these. 40% of the marks are AO1, for applying a standard technique correctly, and the remaining 60% are for reasoning and for solving problems — so an answer with no working rarely scores everything that was available. The bulk of it sits comfortably on the non-calculator paper; a calculator quietly hides the gap until the exam.
- A1 — Algebraic notation
- A2 — Substitution into formulae
- A3 — Expressions, equations, formulae, identities and inequalities
- A4 — Simplifying, expanding and factorising (part Higher)
- A5 — Rearranging formulae
- A6 — Identities, equivalence and algebraic proof (part Higher)
- A7 — Functions, inverse and composite functions (part Higher)
- A8 — Coordinates in all four quadrants
- A9 — Straight-line graphs and y = mx + c (part Higher)
- A10 — Gradients and intercepts of linear functions
- …and 15 more algebra statements at this tier
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Algebra at GCSE Higher: the 25 DfE statements
Every question on this site is filed against one of these statements. Pick one to practise it on its own.
Sample algebra questions for GCSE Higher
- Factorise fully 5x + 5y − 5(a)5(xy − 1)(b)5(x + y + 1)(c)5(x + y − 1)(d)5(x + y − 5)
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5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising. - Factorise fully 56x − 24(a)8(7x − 16)(b)8(56x − 3)(c)8(7x + 3)(d)8(7x − 3)
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8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8. - Expand and simplify (x − 3)²(a)x² + 6x + 9(b)x² − 9(c)x² − 3x + 9(d)x² − 6x + 9
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x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3). - Simplify (5x² + 3x − 2) − (2x² − x + 5)(a)3x² + 4x + 3(b)3x² + 2x + 3(c)7x² + 2x + 3(d)3x² + 4x − 7
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3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3. - Expand and simplify 4(2x − 1) − 3(x + 2) − 5x(a)−10(b)5x − 10(c)2(d)−7
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−10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
Frequently asked questions
How much of the GCSE Higher paper is algebra?
30% of the total marks, and the three papers each carry an equal share of the qualification, so it is spread across all of them rather than concentrated in one.
How many topics are there in algebra at GCSE Higher?
25 DfE content statements, coded A1 to A25 within this area. Every question in the bank is tagged to one of them.
Which algebra topics are Higher only?
Translations and reflections of graphs (A13), Gradients and areas under graphs (A15), Equation of a circle and its tangent (A16) and Iteration (A20). A further 12 statements have a Higher-only part inside otherwise shared content.
Can I use a calculator for algebra questions?
Paper 1 is non-calculator; Papers 2 and 3 allow one. Of the 25 statements in this area, 20 are naturally non-calculator and 1 naturally calculator, with the rest workable either way — so practise both.
What kind of marks does algebra carry?
The GCSE Higher split is 40% AO1 (use and apply standard techniques), 30% AO2 (reason, interpret and communicate) and 30% AO3 (solve problems in and out of context). Questions in this area are set across all three.
Are these past paper questions?
No. Every question at /gcse-higher is original, written to the DfE content statements and checked before it is published. We do not host past papers or mark schemes.
Do I need an account?
No — you can start practising straight away. Progress is saved automatically in your browser.
Is it free?
Yes. The questions, worked answers and printable worksheets are all free, with no adverts.
Tips for algebra at GCSE Higher
- Use the quadratic formula x = (−b ± √(b² − 4ac)) ÷ 2a when a quadratic does not factorise, and read the discriminant b² − 4ac first: positive means two roots, zero means one repeated root, negative means no real roots.
- Completing the square rewrites x² + bx + c as (x + b/2)² − (b/2)² + c. The turning point of the graph is then read straight off: y = (x − p)² + q has its minimum at (p, q).
- Algebraic fractions obey the same rules as numerical fractions. To simplify, factorise the numerator and the denominator first, then cancel a whole common factor — never a single term out of a sum.
- For a quadratic inequality, find where the expression equals zero, sketch the parabola, and read off the interval. Multiplying or dividing an inequality by a negative number reverses the sign.
- For a linear and a quadratic equation together, make y (or x) the subject of the linear one and substitute into the quadratic. Aim for a single quadratic in one unknown, then substitute each root back to find its partner.
Common mistakes (and how to avoid them)
- Expanding (x + 3)² as x² + 9 instead of x² + 6x + 9.
- Cancelling a term rather than a factor: simplifying (x + 4) ÷ 4 to x + 1. Only a factor common to the whole numerator and the whole denominator can be cancelled.
- A sign error in the quadratic formula — forgetting that b appears as −b, so when b is negative the sign flips.
- Solving a quadratic inequality like a linear one and giving a single boundary instead of an interval or two intervals.
Worked examples
- x² − 9 is a difference of two squares: (x − 3)(x + 3). Cancel the common factor (x + 3), for x ≠ −3, leaving x − 3.
- The common denominator of 1/x and 2/(x + 1) is x(x + 1).
- 1/x = (x + 1) ÷ [x(x + 1)] and 2/(x + 1) = 2x ÷ [x(x + 1)].
- Add the numerators: (x + 1 + 2x) ÷ [x(x + 1)] = (3x + 1) ÷ [x(x + 1)].
- The expression is zero at x = 1 and x = −4.
- It is a parabola opening upwards (the x² coefficient is positive), so it is negative between the roots.
- The product is negative for −4 < x < 1.
- Both expressions equal y, so x² − 2x = x + 4.
- Rearrange: x² − 3x − 4 = 0.
- Factorise: (x − 4)(x + 1) = 0, so x = 4 or x = −1.
- Substitute into y = x + 4: x = 4 gives y = 8; x = −1 gives y = 3.
Algebra is 30% of the Higher paper — the largest single area — and every other topic leans on it. Confidence with the quadratic formula, completing the square, algebraic fractions and inequalities saves you expensive slips right across the exam. Always check a solution in the original equation, especially after squaring or clearing a denominator, and write the working: method marks are real marks. Practise a wide range until the steps are automatic, so that in the exam your attention goes on the thinking, not the technique.
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