amazon.co.ukIndex laws, surds, expanding and factorising, algebraic fractions, polynomial division and the factor theorem - A-Level pure maths algebra (Edexcel 9MA0 Topic 2, AQA 7357 Section B) revised with worked examples, practice questions and mark-scheme logic.
Almost every pure mathematics paper opens with an algebraic manipulation worth 3-6 marks, and those marks are lost through slips, not through difficulty. The index laws (Edexcel 9MA0 Topics 2.1-2.2; AQA 7357 B1-B2; OCR A H640 2.01; WJEC/Eduqas and CCEA AS Pure Mathematics) are non-negotiable: x^a x x^b = x^(a+b), x^a / x^b = x^(a-b), (x^a)^b = x^(ab), x^0 = 1 and x^(-a) = 1/x^a. Negative indices send the power downstairs; fractional indices convert to roots, so x^(1/2) = sqrt(x), x^(1/3) = cuberoot(x) and, in general, x^(m/n) = (nth root of x)^m. The A* habit is to convert everything to index form before combining, rather than juggling roots and powers at the same time.
Surds are irrational roots kept in exact form. Two results do all the work: sqrt(a) x sqrt(b) = sqrt(ab) and sqrt(a)/sqrt(b) = sqrt(a/b). To rationalise a denominator of the form a + sqrt(b), multiply top and bottom by the conjugate a - sqrt(b), because (a + sqrt(b))(a - sqrt(b)) = a^2 - b, which is rational. Papers routinely test surds and indices together.
Simplify (4x^2 y^(-1))^(3/2) / (2x^3 y)^(1/2), giving the answer with positive indices.
Line 1: Numerator: (4x^2 y^(-1))^(3/2) = 4^(3/2) x^3 y^(-3/2) = 8x^3 y^(-3/2), using (x^a)^b = x^(ab) and 4^(3/2) = (sqrt(4))^3 = 8.
Line 2: Denominator: (2x^3 y)^(1/2) = 2^(1/2) x^(3/2) y^(1/2).
Line 3: Divide: (8 / 2^(1/2)) x^(3 - 3/2) y^(-3/2 - 1/2) = (8/sqrt(2)) x^(3/2) y^(-2).
Line 4: Rationalise 8/sqrt(2) = 8sqrt(2)/2 = 4sqrt(2), and flip y^(-2): answer 4sqrt(2) x^(3/2) / y^2.
Expanding uses the distributive law: every term inside the bracket is multiplied by the term outside. For double brackets, FOIL (first, outer, inner, last) guarantees no pair is missed: (3x - 2)(x + 4) = 3x^2 + 12x - 2x - 8 = 3x^2 + 10x - 8. Three identities should be automatic rather than re-derived: the difference of two squares a^2 - b^2 = (a + b)(a - b), and the perfect squares (a + b)^2 = a^2 + 2ab + b^2 and (a - b)^2 = a^2 - 2ab + b^2.
The classic mark-losing errors are all sign or structure slips: writing (x + 3)^2 = x^2 + 9 (forgetting 2ab), mishandling a negative multiplier so that -2(x - 1)^2 becomes 2x^2 - 4x + 2 instead of -2x^2 + 4x - 2, and failing to spot differences of two squares in disguise, such as (x - sqrt(3))(x + sqrt(3)) = x^2 - 3 or (2x - 5)(2x + 5) = 4x^2 - 25.
Factorising reverses the process. When a = 1, find two numbers with product c and sum b. When a is not 1, use the AC method: for 6x^2 + 13x + 6, compute ac = 36 and look for two numbers with product 36 and sum 13, namely 9 and 4, so 6x^2 + 13x + 6 = 6x^2 + 9x + 4x + 6 = 3x(2x + 3) + 2(2x + 3) = (2x + 3)(3x + 2). Factorising by grouping is the same pairing idea applied to four terms: x^3 + 2x^2 + 3x + 6 = x^2(x + 2) + 3(x + 2) = (x + 2)(x^2 + 3). Always finish by factorising completely: (x^2 - 4)(x + 5) is not fully factorised until written (x - 2)(x + 2)(x + 5).
Expand and simplify (3x - 2)(x + 4) - (x - 1)^2.
Line 1: (3x - 2)(x + 4) = 3x^2 + 12x - 2x - 8 = 3x^2 + 10x - 8.
Line 2: (x - 1)^2 = x^2 - 2x + 1, so - (x - 1)^2 = -x^2 + 2x - 1.
Line 3: Subtract: (3x^2 + 10x - 8) - (x^2 - 2x + 1) = 2x^2 + 12x - 9.
The single governing rule: cancel factors, never terms. In (x^2 + 3x)/x^2, the numerator factorises to x(x + 3), the x cancels with one x downstairs, and the fraction simplifies to (x + 3)/x. But 3 in (x + 3)/x^2 is a term, not a factor, so nothing cancels. The instant check: substitute x = 1 into the original and the simplified form - if the values differ, an illegal cancellation has happened.
To multiply fractions, factorise everything first, then multiply numerators and denominators and cancel across the whole product. To divide, multiply by the reciprocal of the second fraction. To add or subtract, rewrite both fractions over a common denominator before combining numerators, then factorise the result to see whether it cancels.
Simplify (x^3 - 8)/(x^2 - 4).
Line 1: x^3 - 8 is a difference of two cubes: x^3 - 8 = (x - 2)(x^2 + 2x + 4).
Line 2: x^2 - 4 = (x - 2)(x + 2).
Line 3: Cancel the common factor (x - 2): (x^3 - 8)/(x^2 - 4) = (x^2 + 2x + 4)/(x + 2), for x is not 2 and x is not -2.
Write 3/(x - 1) + 2/(x + 2) as a single fraction.
Line 1: Common denominator: (x - 1)(x + 2).
Line 2: 3(x + 2)/((x - 1)(x + 2)) + 2(x - 1)/((x - 1)(x + 2)) = (3x + 6 + 2x - 2)/((x - 1)(x + 2)).
Line 3: Combine like terms: (5x + 4)/((x - 1)(x + 2)). The numerator has no common factor with the denominator, so this is fully simplified.
These skills feed directly into partial fractions in Year 2 (Edexcel 9MA0 2.5; AQA 7357 B8), where a proper rational function is split into simpler fractions before integration - and partial fraction questions are marked almost entirely on correct factorisation and common-denominator work.
Long division writes dividend = divisor x quotient + remainder. Divide term by term in descending powers: to divide 2x^3 - 5x^2 + 3x - 1 by x - 2, ask what 2x^3 / x is (2x^2), multiply back, subtract, and repeat. The result is 2x^2 - x + 1 with remainder 1, i.e. 2x^3 - 5x^2 + 3x - 1 = (x - 2)(2x^2 - x + 1) + 1.
The factor theorem (Edexcel 9MA0 2.5; AQA 7357 B4; OCR A H640 6.01; CCEA AS1/A21) converts division into substitution: f(a) = 0 if and only if (x - a) is a factor. It is used in two ways. First, to show a given linear expression is a factor - evaluate and show zero. Second, to find unknown coefficients: if (x - 3) is a factor of x^3 - 4x^2 + kx + 6, then f(3) = 27 - 36 + 3k + 6 = 0, so 3k = 3 and k = 1.
The remainder theorem finds the remainder without dividing: the remainder when f(x) is divided by (x - a) is f(a), and when divided by (ax - b) it is f(b/a). For the division above, f(2) = 2(8) - 5(4) + 3(2) - 1 = 1, confirming the remainder - a two-line check worth doing in the exam.
Given f(x) = 2x^3 - 3x^2 - 4x + 4, show that (x - 2) is a factor and factorise f(x) completely.
Line 1: f(2) = 2(8) - 3(4) - 4(2) + 4 = 16 - 12 - 8 + 4 = 0, so (x - 2) is a factor by the factor theorem.
Line 2: Divide: f(x) = (x - 2)(2x^2 + x - 2), obtained by long division or by matching coefficients.
Line 3: Factorise the quadratic by the AC method: 2x^2 + x - 2 = (2x - 1)(x + 1), since ac = -4 and the pair 2, -1 has product -4 and sum 1.
Line 4: f(x) = (x - 2)(2x - 1)(x + 1) - three linear factors, fully factorised.
Once a cubic is factorised, its roots and its sketch follow immediately, and equating coefficients (comparing x^2, x and constant terms on both sides of an identity) is the algebraic alternative to division that examiners accept fully.
This topic is the opening question of pure papers on every board: typically questions 1-3 of Edexcel 9MA0 Paper 1 or Paper 2, AQA 7357 Paper 1, OCR A H640 Paper 1, WJEC/Eduqas Unit 1 and CCEA AS1/A21, worth 3-8 marks each. Command words are precise. Expand and simplify demands one final collected expression. Factorise completely demands every bracket linear (or irreducible quadratic), so check for a hidden difference of two squares. Show that demands every line of working be visible - do not skip the substitution step in a factor theorem argument. Hence means the previous result must be used.
Five A* habits: (1) give exact answers with surds and fractions unevaluated unless a decimal is demanded; (2) convert roots to fractional indices before combining; (3) write a minus sign before a bracket and then distribute it - subtracting a bracket is where most marks die; (4) after any cancellation, substitute x = 1 into original and simplified forms to verify; (5) when factorising, state the zero substitutions that justify each factor, since many mark schemes award the method mark for f(a) = 0 alone.
Typical whole-paper chains: an index simplification (2-3 marks) leading into solving an equation like 9^x = 3^(x + 1); or a factor theorem question feeding a coordinate geometry or integration part later in the same paper. Practising the manipulation with the follow-on part trains the fluency the papers reward: e.g. if (x - 2) is a factor of f(x) = x^3 + ax^2 + bx - 8 and the remainder when f(x) is divided by (x - 1) is -12, then f(2) = 4a + 2b = 0 and f(1) = a + b - 9 = -12, giving a + b = -3 and 2a + b = 0, so a = 3, b = -6. Two simultaneous equations from two theorem substitutions is a standard A* pattern.