AQA A-Level Maths: Advanced Algebra — In-Depth Revision Guide (7357)

Advanced algebra is the engine room of AQA A-Level Maths (7357). Section B of the specification — Algebra and Functions — is where the manipulation skills sit that every other topic depends on. Calculus, coordinate geometry, trigonometric proofs, the binomial theorem and even the applied content all assume that you can rearrange, factorise, simplify and reason with algebraic expressions at speed. If your algebra is sharp, the rest of A-Level Maths becomes far more approachable. If it is shaky, even straightforward calculus questions become a slog.

This guide is the in-depth companion for AQA's algebra and functions content. This is the depth course — combining Year 1 and Year 2 algebra into one focused programme. It runs from polynomial manipulation through algebraic inequalities and advanced simultaneous equations, into the functions toolbox (domain and range, composite and inverse functions), the modulus function, graph transformations, partial fractions in depth, advanced binomial expansion, and factor-remainder applications. Each section gives the core skills, the typical pitfalls, a worked example, and a link to the matching lesson.

The aim is not to replace working through problems. The only way to get good at algebra is to do hundreds of questions until the patterns become automatic. The aim of this guide is to give you a clear map of what AQA can examine, in the order they teach it, so your revision is targeted rather than scattered.

What the AQA 7357 Specification Covers

The AQA A-Level Maths qualification (7357) is assessed through three two-hour papers, each worth 100 marks. Paper 1 is pure, Paper 2 covers pure and mechanics, and Paper 3 covers pure and statistics. Section B of the specification covers Algebra and Functions, and its content can appear on any of the three papers — including inside applied questions, where the algebra is often the bottleneck rather than the modelling.

The table below shows the depth-course sub-topics, the part of the specification they sit under, and a realistic estimate of how many marks across a paper sitting come from each area.

Topic	Spec Reference	Typical paper marks weight
Polynomial manipulation	B1	4-6 marks
Algebraic inequalities	B5	4-6 marks
Advanced simultaneous equations	B4	4-6 marks
Functions: domain and range	B7	3-5 marks
Composite and inverse functions	B7	4-6 marks
Modulus function	B9	4-6 marks
Graph transformations	B8	4-6 marks
Partial fractions in depth	B6	4-6 marks
Advanced binomial expansion	E4	6-8 marks
Factor and remainder applications	B2	4-6 marks

These weights are estimates based on the spread of typical 7357 papers — not guarantees for any single year. What is reliable is that algebra and functions is consistently the most heavily examined area of pure across the three papers, and that the same skills resurface inside calculus, trigonometry and binomial questions.

Polynomial Manipulation

A polynomial is an expression of the form $a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$an​xn+an−1​xn−1+…+a1​x+a0​, where the powers are non-negative integers and the coefficients are real. The degree is the highest power of $x$x that appears. AQA expects fluency with polynomials up to degree 4, including expansion, collection, division and factorisation.

The core operations are: expanding products of polynomials, simplifying by collecting like terms, polynomial division (long division or by inspection), and factorising using known roots. You also need to recognise standard expansion patterns at sight — the difference of two squares $a^2 - b^2 = (a-b)(a+b)$a2−b2=(a−b)(a+b), the sum and difference of two cubes $a^3 \pm b^3$a3±b3, and the binomial expansions of $(a+b)^2$(a+b)2 and $(a+b)^3$(a+b)3.

End behaviour and multiplicity are the two sketching tools to internalise. The end behaviour is determined by the leading term: a positive leading coefficient with even degree rises on both sides; a positive leading coefficient with odd degree falls on the left and rises on the right. Multiplicity describes how the curve meets the x-axis at each root. A simple root means the curve crosses cleanly; a double root means it touches and turns; a triple root means it flattens through the axis.

A short worked example. Expand $(x-1)(x+2)(x-3)$(x−1)(x+2)(x−3). Multiply the first two: $(x-1)(x+2) = x^2 + x - 2$(x−1)(x+2)=x2+x−2. Multiply by $(x-3)$(x−3): $x^3 - 2x^2 - 5x + 6$x3−2x2−5x+6. The cubic has roots at $x = 1, -2, 3$x=1,−2,3 by construction, and rises on the right.

A common pitfall is sign-error chains during long polynomial division. Another is treating a repeated root as two intersections rather than a single touch point. For full coverage, see the Polynomial Manipulation lesson.

Algebraic Inequalities

Inequalities at A-Level extend GCSE work into three connected areas: linear inequalities, quadratic inequalities, and rational inequalities. The mechanics look similar to equations, but the reasoning is different and a single careless step can flip the answer.

For linear inequalities, the only rule that matters is that multiplying or dividing both sides by a negative number reverses the inequality sign. This is the single most common source of lost marks at A-Level — many candidates know the rule but forget it under exam pressure.

For quadratic inequalities of the form $ax^2 + bx + c > 0$ax2+bx+c>0, the right method is graphical. Rearrange so one side is zero, find the roots, sketch the parabola, and read off the region where the curve sits above or below the x-axis as required. Algebraic shortcuts like "divide through by the unknown sign" are unreliable and usually wrong.

For rational inequalities of the form $\frac{x+1}{x-2} > 0$x−2x+1​>0, do not multiply both sides by $(x-2)$(x−2) because its sign is unknown. Instead, find the critical values where numerator or denominator equals zero, draw a number line, and test the sign of the expression in each interval. The solution is the union of intervals where the expression has the correct sign. Be careful with strict versus non-strict inequalities at the critical values — the denominator can never equal zero, even if the inequality is non-strict.

A worked example. Solve $\frac{x-1}{x+3} \le 0$x+3x−1​≤0. The critical values are $x = 1$x=1 and $x = -3$x=−3 (excluded). Test each interval: for $x < -3$x<−3 positive; for $-3 < x < 1$−3<x<1 negative; for $x > 1$x>1 positive. The solution set is $-3 < x \le 1$−3<x≤1.

For sketching diagrams and a clean sign-table method, see the Algebraic Inequalities lesson.

Advanced Simultaneous Equations

AQA tests several flavours of simultaneous equations: pairs of linear equations, linear-quadratic systems, two-quadratic systems, and three-equation linear systems. The linear pair is GCSE-level, but the rest are genuine A-Level skills and a frequent source of marks.

The standard method for a linear-quadratic pair is substitution. Rearrange the linear equation to make $x$x or $y$y the subject, substitute into the quadratic, and solve. You will end up with a quadratic in one variable, which factorises (or yields to the quadratic formula) to give the x-values of the intersection points. Substitute back to find the y-values.

The discriminant has a powerful geometric meaning here. If a line and a curve are simultaneous and you reduce them to a quadratic, the discriminant of that quadratic tells you about their intersection. Two distinct real roots means the line cuts the curve at two points. A repeated root means the line is tangent to the curve. No real roots means they do not meet. Tangency conditions are a classic AQA question — you are given a parameter $k$k and asked to find the value for which the line is tangent.

For two-quadratic systems such as a circle intersecting a parabola, subtract one equation from the other to eliminate the quadratic term and produce a linear equation in $x$x and $y$y. This linear equation is the chord through the intersection points. Substitute it into either original equation to finish.

A common pitfall is solving for $x$x and forgetting to find the corresponding y-values, or finding only one intersection when the question asks for two. Take your time with the algebra; rushing here costs more marks than rushing anywhere else.

For worked tangency examples, see the Advanced Simultaneous Equations lesson.

Functions, Domain and Range

A function is a rule that assigns to each input from a set called the domain exactly one output, and the set of all outputs is called the range. AQA requires you to be precise with this language: A-Level function questions that fail full marks usually fail on domain or range, not on the algebra inside.

The domain is the set of inputs for which the function is defined. For a real-valued function, you exclude inputs that produce division by zero, even roots of negative numbers, or logarithms of non-positive numbers. So $f(x) = \sqrt{x-2}$f(x)=x−2​ has domain $x \ge 2$x≥2, and $g(x) = \frac{1}{x-3}$g(x)=x−31​ has domain $x \ne 3$x=3.

To find a range, sketch the function and read the y-values it covers. For a quadratic with positive leading coefficient and minimum at $(h,k)$(h,k), the range is $f(x) \ge k$f(x)≥k. For an exponential like $f(x) = e^x + 1$f(x)=ex+1, the range is $f(x) > 1$f(x)>1 because $e^x$ex is always positive but never zero.

A function is one-to-one if no two distinct inputs share an output — equivalently, if any horizontal line crosses the graph at most once. A function is many-to-one if some output is shared by multiple inputs. This distinction matters for inverses: only one-to-one functions have inverses on their full domain. A many-to-one function can be made one-to-one by restricting its domain.

A worked example. State the domain and range of $f(x) = \sqrt{4-x^2}$f(x)=4−x2​. The expression under the root must be non-negative: $4 - x^2 \ge 0$4−x2≥0, giving $-2 \le x \le 2$−2≤x≤2 (the domain). The function is the upper semicircle of radius 2, so the range is $0 \le f(x) \le 2$0≤f(x)≤2.

For full coverage with sketching practice, see the Functions: Domain and Range lesson.

Composite and Inverse Functions

The composite function $fg$fg means "do $g$g first, then $f$f." So $fg(x) = f(g(x))$fg(x)=f(g(x)). The order matters: in general $fg \ne gf$fg=gf. This is the single most common error in this topic — students apply $f$f first because it is written first.

For composites to make sense, the range of $g$g must be a subset of the domain of $f$f. If $g$g produces an output that lies outside the domain of $f$f, the composite is undefined at that input.

The inverse function $f^{-1}$f−1 undoes the action of $f$f, so $f^{-1}(f(x)) = x$f−1(f(x))=x. The domain of $f^{-1}$f−1 is the range of $f$f, and the range of $f^{-1}$f−1 is the domain of $f$f. The graph of $y = f^{-1}(x)$y=f−1(x) is the reflection of $y = f(x)$y=f(x) in the line $y = x$y=x.

To find an inverse: write $y = f(x)$y=f(x), swap $x$x and $y$y, and solve for $y$y. For $f(x) = 2x + 3$f(x)=2x+3, swapping gives $x = 2y + 3$x=2y+3, so $f^{-1}(x) = \frac{x-3}{2}$f−1(x)=2x−3​. Always state the domain of the inverse explicitly — it is the range of the original.

A common pitfall is forgetting that only one-to-one functions have inverses. For a quadratic $f(x) = x^2 + 1$f(x)=x2+1, you cannot invert on $\mathbb{R}$R because it is many-to-one. AQA usually fixes this by giving a restricted domain such as $x \ge 0$x≥0. The inverse is then $f^{-1}(x) = \sqrt{x-1}$f−1(x)=x−1​ with domain $x \ge 1$x≥1.

For composite domain checking and a clean inverse workflow, see the Composite and Inverse Functions lesson.

The Modulus Function

The modulus function $|x|$∣x∣ returns the magnitude of $x$x, ignoring sign. So $|x| = x$∣x∣=x when $x \ge 0$x≥0 and $|x| = -x$∣x∣=−x when $x < 0$x<0. This piecewise definition is the foundation of every modulus problem.

For sketching, distinguish two transformations carefully. $y = |f(x)|$y=∣f(x)∣ takes any part of $y = f(x)$y=f(x) that lies below the x-axis and reflects it upward. $y = f(|x|)$y=f(∣x∣) takes the part of $y = f(x)$y=f(x) for $x \ge 0$x≥0 and reflects it across the y-axis. The two look similar at first but produce very different graphs, and AQA tests both.

To solve modulus equations like $|2x - 3| = 5$∣2x−3∣=5, split into cases: either $2x - 3 = 5$2x−3=5 giving $x = 4$x=4, or $2x - 3 = -5$2x−3=−5 giving $x = -1$x=−1. Always check the answers in the original. For modulus equations with a modulus on each side, like $|x + 2| = |2x - 1|$∣x+2∣=∣2x−1∣, square both sides and solve the resulting quadratic.

To solve modulus inequalities like $|x - 4| < 3$∣x−4∣<3, rewrite as a double inequality $-3 < x - 4 < 3$−3<x−4<3, giving $1 < x < 7$1<x<7. For $|x - 4| > 3$∣x−4∣>3, the solution is $x < 1$x<1 or $x > 7$x>7. Sketching the modulus graph alongside the constant value usually settles any uncertainty.

A common pitfall is forgetting that a modulus equation can have zero, one, or two solutions depending on the right-hand side. If the right-hand side is negative, the equation has no solution because $|\ldots|$∣…∣ is always non-negative.

For sketching practice and a case-split workflow, see the Modulus Function lesson.

Graph Transformations

Graph transformations move, stretch, or reflect the graph $y = f(x)$y=f(x) to produce related graphs. AQA expects you to know all the standard transformations and to apply them in any order, including to unfamiliar functions where you cannot see the algebra.

Transformation	Effect on the graph
$y = f(x) + a$y=f(x)+a	Translation by $a$a units in the positive y-direction
$y = f(x + a)$y=f(x+a)	Translation by $a$a units in the negative x-direction
$y = a f(x)$y=af(x)	Vertical stretch by scale factor $a$a
$y = f(ax)$y=f(ax)	Horizontal stretch by scale factor $1/a$1/a
$y = -f(x)$y=−f(x)	Reflection in the x-axis
$y = f(-x)$y=f(−x)	Reflection in the y-axis

The single rule that catches most candidates is the "inside-bracket do the opposite" rule. Anything that happens to $x$x inside the bracket affects the graph in the opposite way to what you might expect. Adding 3 inside the bracket — $y = f(x + 3)$y=f(x+3) — translates the graph 3 units to the left, not the right. Multiplying $x$x by 2 inside the bracket — $y = f(2x)$y=f(2x) — stretches the graph by a factor of $\frac{1}{2}$21​ in the x-direction, not 2. Outside-the-bracket transformations behave intuitively; inside-the-bracket transformations invert.

For combining transformations, the order matters. For $y = 2 f(x + 1) - 3$y=2f(x+1)−3 starting from $y = f(x)$y=f(x): first replace $x$x by $x + 1$x+1 (translate left by 1), then multiply by 2 (vertical stretch by 2), then subtract 3 (translate down by 3). Drawing the intermediate graphs is far safer than trying to do it all at once.

A common pitfall is mixing up the direction of horizontal translations or stretches. Another is applying transformations in the wrong order. For step-by-step sketches, see the Graph Transformations lesson.

Partial Fractions in Depth

Partial fractions is the reverse of adding algebraic fractions: starting from a single fraction with a complicated denominator, you split it into a sum of simpler fractions. AQA uses partial fractions in three places — algebra, the binomial expansion of $(1 + x)^n$(1+x)n where $n$n is not a positive integer, and integration — so the technique is high-value across the course.

There are three cases on the 7357 specification.

Case	Denominator factor	Partial fraction form
Distinct linear factors	$(ax + b)(cx + d)$(ax+b)(cx+d)	$\frac{A}{ax+b} + \frac{B}{cx+d}$ax+bA​+cx+dB​
Repeated linear factor	$(ax + b)^2$(ax+b)2	$\frac{A}{ax+b} + \frac{B}{(ax+b)^2}$ax+bA​+(ax+b)2B​
Improper fraction	numerator degree $\ge$≥ denominator degree	polynomial part + proper fraction

In each case, write the original fraction as a sum of unknown-coefficient fractions, multiply both sides by the original denominator, and solve for the unknowns. The two standard solving methods are: substituting strategic values of $x$x to make terms vanish (the cover-up method for distinct linear factors), and equating coefficients of like powers.

The cover-up method is fast. To find $A$A in $\frac{A}{x-1} + \frac{B}{x+2} = \frac{3x+4}{(x-1)(x+2)}$x−1A​+x+2B​=(x−1)(x+2)3x+4​, cover the $(x-1)$(x−1) on the right and substitute $x = 1$x=1: $A = \frac{3 + 4}{1 + 2} = \frac{7}{3}$A=1+23+4​=37​. Repeat for $B$B by covering $(x+2)$(x+2) and substituting $x = -2$x=−2, giving $B = \frac{2}{3}$B=32​.

For an improper fraction where the numerator has degree at least the denominator, divide first. $\frac{x^2 + 3x + 1}{x + 1}$x+1x2+3x+1​ becomes $x + 2 - \frac{1}{x+1}$x+2−x+11​ after polynomial division.

A common pitfall is using the wrong template — for instance, writing $\frac{Ax+B}{(x-1)^2}$(x−1)2Ax+B​ instead of $\frac{A}{x-1} + \frac{B}{(x-1)^2}$x−1A​+(x−1)2B​ for a repeated linear factor. Always factorise the denominator fully before deciding which case applies.

For worked examples in all three cases, see the Partial Fractions in Depth lesson.

Advanced Binomial Expansion

The binomial theorem for a positive integer $n$n gives

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \ldots + x^n.$(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3+…+xn.

The series terminates after $n+1$n+1 terms because the coefficients eventually multiply by zero. This is the Year 1 form, and it works for any real $x$x.

AQA's advanced binomial work extends this to fractional and negative $n$n. The same formula

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \ldots$(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3+…

still holds — but it now produces an infinite series because the coefficients never multiply by zero. Crucially, the series only converges (gives a meaningful value) for $|x| < 1$∣x∣<1. Stating the range of validity is part of the question, and forgetting it costs an easy mark every series.

To expand $(a + bx)^n$(a+bx)n with non-integer $n$n, you must first factor out $a$a to put the bracket in the form $(1 + \text{something})^n$(1+something)n. So $(2 + x)^{-1} = 2^{-1}(1 + \frac{x}{2})^{-1}$(2+x)−1=2−1(1+2x​)−1, and the expansion proceeds with $\frac{x}{2}$2x​ in place of $x$x in the standard formula. The validity is then $\left|\frac{x}{2}\right| < 1$​2x​​<1, i.e. $|x| < 2$∣x∣<2.

The binomial expansion combines beautifully with partial fractions. To expand $\frac{1}{(1-x)(1+2x)}$(1−x)(1+2x)1​ as a series, first split into partial fractions: $\frac{1}{(1-x)(1+2x)} = \frac{A}{1-x} + \frac{B}{1+2x}$(1−x)(1+2x)1​=1−xA​+1+2xB​, find the coefficients, then expand each piece using the binomial series. The validity is the intersection of the individual ranges of validity — the strictest of the conditions wins.

A short worked example. Expand $(1 + 2x)^{-2}$(1+2x)−2 up to the $x^3$x3 term. With $n = -2$n=−2 and $u = 2x$u=2x: constant 1; $x$x term $(-2)(2x) = -4x$(−2)(2x)=−4x; $x^2$x2 term $\frac{(-2)(-3)}{2!}(2x)^2 = 12x^2$2!(−2)(−3)​(2x)2=12x2; $x^3$x3 term $\frac{(-2)(-3)(-4)}{3!}(2x)^3 = -32x^3$3!(−2)(−3)(−4)​(2x)3=−32x3. Validity: $|x| < \frac{1}{2}$∣x∣<21​.

A common pitfall is misreading the sign of $n$n, or forgetting to apply the factor $a^n$an when the bracket starts as $(a + bx)^n$(a+bx)n. Another is omitting the validity statement, which is worth a mark on its own. For full coverage, see the Advanced Binomial Expansion lesson.

Factor and Remainder Applications

The factor theorem says: $(x - a)$(x−a) is a factor of $f(x)$f(x) if and only if $f(a) = 0$f(a)=0. The remainder theorem says: when $f(x)$f(x) is divided by $(x - a)$(x−a), the remainder is $f(a)$f(a). These two results are the standard tools for factorising cubics and higher-order polynomials, and for the unknown-coefficient applications AQA likes to set.

The workflow for factorising a cubic is to test small integer values until you find one that gives $f(a) = 0$f(a)=0, then factor out $(x - a)$(x−a) by polynomial division, leaving a quadratic factor. The candidates to test are divisors of the constant term — by the rational root theorem, any rational root must divide it.

The applications AQA examines include: finding unknown coefficients given factor or remainder information; combining both theorems in a single question; and verifying factor structure before integrating or differentiating.

A worked applications example. Given $f(x) = 2x^3 + ax^2 + bx - 6$f(x)=2x3+ax2+bx−6 has $(x-1)$(x−1) as a factor and leaves remainder 12 when divided by $(x+2)$(x+2), find $a$a and $b$b. From $f(1) = 0$f(1)=0: $2 + a + b - 6 = 0$2+a+b−6=0, so $a + b = 4$a+b=4. From $f(-2) = 12$f(−2)=12: $-16 + 4a - 2b - 6 = 12$−16+4a−2b−6=12, so $2a - b = 17$2a−b=17. Adding: $3a = 21$3a=21, so $a = 7$a=7, $b = -3$b=−3.

A common pitfall is sign errors during polynomial division, and failing to check enough small values before moving to the formula. Test $x = 1, -1, 2, -2$x=1,−1,2,−2 first; one is usually a root.

For polynomial division technique and worked applications, see the Factor and Remainder Applications lesson.

Common Mark-Loss Patterns Across Advanced Algebra

Across the whole advanced algebra section, a small set of habits accounts for a disproportionate share of lost marks. None of these are about content you do not know. They are all about content you do know, applied carelessly.

• Sign errors when expanding negative brackets. $-3(x - 2) = -3x + 6$−3(x−2)=−3x+6, not $-3x - 6$−3x−6. Many candidates lose marks here in the final line of an otherwise correct piece of work.
• Cancelling terms instead of factors. A common pitfall is writing $\frac{x+4}{x+2}$x+2x+4​ as 2 — you can only cancel complete bracketed factors that appear in both numerator and denominator.
• Forgetting to flip the inequality when multiplying or dividing by a negative. Slowing down and underlining each negative-multiplication step is a cheap and effective discipline.
• Solving a quadratic and not finding both intersection coordinates in simultaneous-equation problems. The question usually asks for points; an x-value alone is half the answer.
• Stating a domain or range as a one-sided inequality when both sides matter. Especially common with restricted-domain inverses.
• Using the wrong partial-fraction template for repeated linear factors. Always check the case before assigning unknown coefficients.
• Treating modulus equations as automatically two-solution problems without checking which branches actually satisfy the original equation.
• Mis-applying the inside-bracket rule for graph transformations. Inside the bracket, do the opposite.
• Omitting the range of validity on a binomial expansion with non-integer $n$n. This is usually one mark on its own.
• Not showing enough working. AQA mark schemes award method marks generously when the working is clear. A correct final answer with no working can score fewer marks than an incorrect final answer with clean method.

A revision plan that explicitly drills these habits — not just the content — will move your grade more than another pass through the textbook.

Recommended Six-Week Revision Plan

This plan is designed for a candidate who has covered the algebra and functions content in lessons but wants to revise it in depth before the exam. It assumes about 5-6 hours per week on this section. Adjust pace if you are starting earlier or later.

Week	Topics	Practice
1	Polynomial manipulation; algebraic inequalities (linear, quadratic, rational)	30-40 expansion and division questions; 20 inequality problems including 5 rational
2	Advanced simultaneous equations; functions, domain and range	15 simultaneous-equation problems including 5 with tangency; 15 domain/range problems with sketches
3	Composite and inverse functions; modulus function	15 composite-function problems with domain checks; 10 modulus-equation problems; 10 modulus-sketching problems
4	Graph transformations (single and combined); partial fractions in depth (all cases)	15 transformation problems; 15 partial-fraction decompositions covering each case
5	Advanced binomial expansion (including partial-fraction expansion); factor and remainder applications	15 binomial expansions with validity statements; 10 factor-remainder applications problems
6	Mixed practice; targeted review of weakest topics; full algebra-and-functions question sets	One full mixed problem set per day; review marking-scheme working for any question scoring below 60%

The point of the plan is to keep moving forward while maintaining contact with earlier topics. Do not spend three weeks on polynomial manipulation and run out of time before partial fractions or binomial expansion. By the end of week 5, every topic in the depth course should have had focused contact and a practice round. Week 6 is consolidation and weakness-targeting.

A useful discipline through the whole plan is to treat any question you got wrong not as a mistake but as a diagnostic. Was it a content gap? A method error? A careless arithmetic slip? Logging the cause means your next review session targets the right thing.

How LearningBro's AQA Advanced Algebra Course Helps

LearningBro's AQA A-Level Maths: Advanced Algebra course is built around the structure of this guide. Each of the ten lessons covers one focus area of Section B of the 7357 specification — combining Year 1 and Year 2 algebra into one in-depth programme — with worked examples, practice questions and mark-scheme-style solutions. Lessons end with a short review and quick-recall questions designed for spaced revisits.

The course is designed to be used in two ways. As a first pass, you can work through the lessons in order, building each topic on the last. As a revision tool, you can drop into any lesson and work the practice independently — for example, drilling partial fractions and the advanced binomial together for a week before mocks. The AI tutor is available throughout to give targeted hints when you get stuck, without giving away full solutions, and to mark your written working with structured feedback.

If you want one place to revise advanced algebra well, with realistic practice and clean explanations of every topic, the full course is the right next step. Start with the AQA A-Level Maths: Advanced Algebra course.

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• AQA A-Level Maths: Pure 1 Revision Guide
• AQA A-Level Maths: Pure 2 Revision Guide
• AQA A-Level Maths: Paper Strategy Guide