Algebraic Manipulation · Study Notes · Edexcel IGCSE 4MA1 Higher

Expand, simplify, collect like terms. The bedrock of every algebra question — get this fluent and the rest follows.

At a glance

Single bracket: $a(b + c) = ab + ac$ .
Double bracket: FOIL — First, Outer, Inner, Last.
Square: $(a + b)^{2} = a^{2} + 2ab + b^{2}$ (THREE terms).
Difference of squares: $(a-b)(a+b) = a^{2} - b^{2}$ .
Three brackets: do two first, then expand.
Like terms: same variable, same power.
Identity $\equiv$ vs equation $=$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.1 — Expand single, double and multi-brackets.
2.1 — Simplify expressions by collecting like terms.
2.1 — Use the difference of squares pattern.
2.1 — Distinguish expressions, equations, and identities.

Expanding brackets — FOIL and beyond

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Multiply each pair, simplify. Show every step on Edexcel.

Single bracket. $a(b + c) = ab + ac$ . Multiply $a$ by each term inside.

Example: $3x(2x - 5) = 6x^{2} - 15x$ .

Double brackets — FOIL. $(a + b)(c + d) = ac + ad + bc + bd$ .

Step	Description
F	First × First
O	Outer × Outer
I	Inner × Inner
L	Last × Last

FOIL is a 2×2 grid: every term in one bracket times every term in the other.

Example: $(2x + 3)(x - 4)$

F: $2x \cdot x = 2x^{2}$ .
O: $2x \cdot (-4) = -8x$ .
I: $3 \cdot x = 3x$ .
L: $3 \cdot (-4) = -12$ .
Combine: $2x^{2} - 8x + 3x - 12 = 2x^{2} - 5x - 12$ .

Three brackets. Do TWO first, then expand by the third.

Example: $(x + 1)(x - 2)(x + 3)$

$(x+1)(x-2) = x^{2} - x - 2$ .
$(x^{2} - x - 2)(x + 3) = x^{3} + 3x^{2} - x^{2} - 3x - 2x - 6$ .
Simplify: $x^{3} + 2x^{2} - 5x - 6$ .

Squaring binomials — TWO patterns to memorise:

$\boxed{(a + b)^{2} = a^{2} + 2ab + b^{2}}$ $\boxed{(a - b)^{2} = a^{2} - 2ab + b^{2}}$

Both have THREE terms. Forgetting the middle is a classic A* trap.

The two ab rectangles are the middle term 2ab that students often forget.

Difference of squares:

$\boxed{(a - b)(a + b) = a^{2} - b^{2}}$

Two terms — the cross-terms cancel.

Worked qualitative. Why does $(a + b)(a - b)$ give only two terms?

FOIL: $a^{2} - ab + ab - b^{2}$ .
Middle terms $-ab + ab = 0$ .
Result: $a^{2} - b^{2}$ — two terms.
Useful for factoring questions later.

Edexcel tip. Show every FOIL step. Mark schemes give M1 for any 3 of 4 terms correct, A1 for the simplified answer.

Single: $a(b+c) = ab + ac$ .
Double: FOIL.
Square: three terms.
Difference of squares: two terms.
Three brackets: do two first.

Simplifying — collecting like terms

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Same variable, same power = like. Combine coefficients.

Like terms have the same variable raised to the same power.

Like?	Why
$3x$ and $5x$	✓ Same variable, same power
$4x^{2}$ and $-7x^{2}$	✓ Same
$2x$ and $2x^{2}$	✗ Different powers
$3x$ and $3y$	✗ Different variables
$5xy$ and $-2xy$	✓ Same

Collecting: add the coefficients of like terms.

Example: $5x^{2} - 3x + 7 - 2x^{2} + 5x - 4$ .

Group:

$x^{2}$ : $5x^{2} - 2x^{2} = 3x^{2}$ .
$x$ : $-3x + 5x = 2x$ .
Constants: $7 - 4 = 3$ .

Result: $3x^{2} + 2x + 3$ .

Algebra fluency rules:

$x \cdot x = x^{2}$ .
$x \cdot 0 = 0$ .
$0 \cdot x = 0$ .
$1 \cdot x = x$ (drop the $1$ ).
$-1 \cdot x = -x$ .
$x + x = 2x$ (NOT $x^{2}$ ).
$x \cdot x = x^{2}$ (NOT $2x$ ).

Worked qualitative. Why does $x + x = 2x$ but $x \cdot x = x^{2}$ ?

$x + x$ : adding two of the same — count them. $2x$ .
$x \cdot x$ : multiplying two of the same — multiply means power up. $x^{2}$ .
Different operations, different rules.

Edexcel tip. When simplifying multi-power expressions, write powers DESCENDING (highest first): $3x^{2} + 2x + 3$ . Mark schemes prefer this convention.

Like terms: same variable AND same power.
Combine coefficients only.
Order by descending power.
$x + x = 2x$ . $x \cdot x = x^{2}$ .

Expression vs equation vs identity

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Three different objects with three different jobs.

Expression. A combination of numbers, variables, operations — NO equals sign.

Examples: $3x^{2} + 2x - 5$ , $\dfrac{x+1}{x-3}$ , $\sqrt{2x+1}$ .

You can SIMPLIFY an expression but not 'solve' it.

Equation. A statement: two expressions are equal, using $=$ . True for SOME values of the variable (the solutions).

Examples: $3x + 2 = 11$ (solution $x = 3$ ). $x^{2} = 16$ (solutions $x = \pm 4$ ).

You SOLVE equations.

Identity. A statement that is TRUE for ALL values of the variable. Often written with $\equiv$ .

Examples:

$(a + b)^{2} \equiv a^{2} + 2ab + b^{2}$ (always true).
$\sin^{2}\theta + \cos^{2}\theta \equiv 1$ (always true).
$a(b + c) \equiv ab + ac$ (distributive identity).

Worked qualitative. Is $x^{2} = 16$ an equation or identity?

True only for $x = 4$ or $x = -4$ , not for ALL $x$ .
It's an EQUATION.

Worked qualitative. Is $(a + b)^{2} = a^{2} + b^{2}$ an identity?

$(1 + 2)^{2} = 9$ , but $1^{2} + 2^{2} = 5$ . Different.
NOT true for all values.
NOT an identity. (And, as a matter of fact, it's never true except when $a = 0$ or $b = 0$ .)

Edexcel tip. Sometimes Edexcel asks 'Show that ... is an identity'. You need to prove it works for ALL values, usually by expanding both sides until they're equal.

Expression: simplify only.
Equation: solve.
Identity: always true; prove by expanding.
$=$ for equations, $\equiv$ for identities.

Quick recap

Single bracket: $a(b+c) = ab + ac$ .
Double: FOIL.
$(a+b)^{2} = a^{2} + 2ab + b^{2}$ — three terms.
Difference of squares: $(a-b)(a+b) = a^{2} - b^{2}$ .
Three brackets: do two first.
Like terms: same variable + same power.
Expression vs equation vs identity.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$(a + b)^{2} = a^{2} + 2ab + b^{2}$
$(a - b)^{2} = a^{2} - 2ab + b^{2}$
$(a + b)(a - b) = a^{2} - b^{2}$
FOIL: First, Outer, Inner, Last.
Like terms: same variable, same power.

How it’s examined

Algebraic manipulation appears every Paper 1H AND 2H, often as the first 2-4 marks of an algebra question, but also as multi-step questions for 4-6 marks. Examiner reports flag (1) writing $(a+b)^{2}$ as $a^{2} + b^{2}$ , (2) sign errors with negative brackets, (3) combining unlike terms.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on algebraic manipulation, written exactly the way a tutor would explain them at the board.

1Expand a single bracket
Foundation• expand
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Question
Expand $3x(2x^{2} - 5x + 4)$ .
Step-by-step solution
1. Step 1
  Multiply each term in the bracket by $3x$ .
  $3x \cdot 2x^{2} = 6x^{3},\ 3x \cdot (-5x) = -15x^{2},\ 3x \cdot 4 = 12x$
2. Step 2
  Combine:
  $3x(2x^{2} - 5x + 4) = 6x^{3} - 15x^{2} + 12x$
Answer
$6x^{3} - 15x^{2} + 12x$
Examiner tip
Use the index laws when multiplying: $3x \cdot 2x^{2} = 6x^{3}$ (multiply numbers, ADD indices). Mark schemes deduct for indexing errors.
2Expand a double bracket
Foundation• Adapted from 4MA1/1H May/Jun 2024 Q5• expand, FOIL
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Question
Expand and simplify $(2x + 3)(x - 4)$ .
Step-by-step solution
1. Step 1
  Use FOIL: First, Outer, Inner, Last.
  $(2x)(x) + (2x)(-4) + (3)(x) + (3)(-4)$
2. Step 2
  Compute each: $2x^{2} - 8x + 3x - 12$ .
3. Step 3
  Combine like terms: $2x^{2} - 5x - 12$ .
Answer
$2x^{2} - 5x - 12$
Examiner tip
Show every FOIL step. Mark scheme awards M1 for any 3 of 4 terms correct, A1 for the simplified answer.
3Square a binomial
Higher• expand, squaring
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Question
Expand $(3x - 2)^{2}$ .
Step-by-step solution
1. Step 1
  $(a - b)^{2} = a^{2} - 2ab + b^{2}$ .
2. Step 2
  $a = 3x$ , $b = 2$ . Substitute:
  $(3x)^{2} - 2(3x)(2) + 2^{2} = 9x^{2} - 12x + 4$
Answer
$9x^{2} - 12x + 4$
Examiner tip
Common error: writing $(3x - 2)^{2}$ as $9x^{2} - 4$ (missing middle term). The expansion has THREE terms, not two.
4Expand three brackets
Higher• Adapted from 4MA1/2H Jan 2024 Q6• expand, three brackets
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Question
Expand and simplify $(x + 1)(x - 2)(x + 3)$ .
Step-by-step solution
1. Step 1
  Multiply the first two brackets first using FOIL: $(x+1)(x-2) = x^{2} - x - 2$ .
2. Step 2
  Multiply the result by $(x + 3)$ :
  $(x^{2} - x - 2)(x + 3) = x^{3} + 3x^{2} - x^{2} - 3x - 2x - 6$
3. Step 3
  Simplify: $x^{3} + 2x^{2} - 5x - 6$ .
Answer
$x^{3} + 2x^{2} - 5x - 6$
Examiner tip
Always do TWO brackets first, then expand by the third. Mark schemes credit each stage; trying all three at once invites errors.
5Simplify by collecting like terms
Foundation• simplify, like terms
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Question
Simplify $5x^{2} - 3x + 7 - 2x^{2} + 5x - 4$ .
Step-by-step solution
1. Step 1
  Group like terms by power.
2. Step 2
  $x^{2}$ : $5x^{2} - 2x^{2} = 3x^{2}$ .
3. Step 3
  $x$ : $-3x + 5x = 2x$ .
4. Step 4
  Constants: $7 - 4 = 3$ .
5. Step 5
  Result: $3x^{2} + 2x + 3$ .
Answer
$3x^{2} + 2x + 3$
Examiner tip
Like terms have the SAME variable to the SAME power. $x^{2}$ and $x$ are NOT like terms. Always sort by power before combining.
6Writing an expression from a description [1 mark]
Getting started• Paper 1H/2H style — 1 mark• symbols, forming expressions
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Question
Pens cost $35$ pence each. Pencils cost $20$ pence each.

Sara buys $n$ pens and $m$ pencils.

Write down an expression, in terms of $n$ and $m$ , for the total cost in pence. [1]
Step-by-step solution
1. Step 1
  The cost of $n$ pens is $35 \times n = 35n$ pence, and the cost of $m$ pencils is $20m$ pence.
2. Step 2
  B1 Add the two costs.
  $35n + 20m$
Answer
$35n + 20m$
Examiner tip
Examiner tip: Test your expression with small numbers. Two pens and one pencil cost $70 + 20 = 90$ pence, and $35(2) + 20(1) = 90$ .

Common mistake: Writing $55nm$ or $35 + 20 + n + m$ . Different letters are not like terms, so $35n$ and $20m$ cannot be combined.
7Simplifying a quotient where a letter ends up in the denominator [2 marks]
Getting started• Paper 1H/2H style — 2 marks• simplifying, quotients of terms
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Question
Simplify fully $\dfrac{15x^{3}y^{2}}{5xy^{4}}$ [2]
Step-by-step solution
1. Step 1
  B1 Cancel the numbers and the powers of $x$ : $15 \div 5 = 3$ and $x^{3} \div x = x^{2}$ .
  $\frac{3x^{2}y^{2}}{y^{4}}$
2. Step 2
  B1 There are more $y$ s on the bottom, so two of them are left in the denominator: $y^{2} \div y^{4} = \dfrac{1}{y^{2}}$ .
  $\frac{3x^{2}}{y^{2}}$
Answer
$\dfrac{3x^{2}}{y^{2}}$ (or $3x^{2}y^{-2}$ )
Examiner tip
Examiner tip: Cancel one letter at a time. Where the larger power is on the bottom, the letter stays on the bottom.

Common mistake: Writing $3x^{2}y^{2}$ by subtracting the powers the wrong way round, or $3x^{3}y^{-2}$ by forgetting to cancel the $x$ . Each correct part earns one B mark, so get as many right as you can.
8Multiplying a bracket by a three-term bracket [2 marks]
Building confidence• Paper 1H/2H style — 2 marks• expanding brackets, collecting like terms
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Question
Expand and simplify $(x + 4)(x^{2} - 3x + 2)$ [2]
Step-by-step solution
1. Step 1
  M1 Multiply each of the three terms in the second bracket by $x$ , then by $4$ , giving six terms (at least four correct).
  $x^{3} - 3x^{2} + 2x + 4x^{2} - 12x + 8$
2. Step 2
  A1 Collect the $x^{2}$ terms and the $x$ terms.
  $x^{3} + x^{2} - 10x + 8$
Answer
$x^{3} + x^{2} - 10x + 8$
Examiner tip
Examiner tip: Two terms times three terms makes six products. Count them before you collect.

Common mistake: Collecting $-3x^{2} + 4x^{2}$ as $-x^{2}$ , or $2x - 12x$ as $-14x$ . Check by substituting $x = 1$ : $(5)(0) = 0$ and $1 + 1 - 10 + 8 = 0$ .
9How much larger one area is than another [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• expanding brackets, area, show that
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Question
Rectangle A has length $(2x + 5)$ cm and width $(x + 2)$ cm.

Rectangle B has length $(x + 4)$ cm and width $(2x - 1)$ cm.

Show that the area of rectangle A is $2(x + 7)$ cm² more than the area of rectangle B. [3]
Step-by-step solution
1. Step 1
  M1 Expand the area of rectangle A, or of rectangle B.
  $(2x + 5)(x + 2) = 2x^{2} + 4x + 5x + 10 = 2x^{2} + 9x + 10$
2. Step 2
  M1 Expand the area of rectangle B and subtract it as a whole, in a bracket.
  $(x + 4)(2x - 1) = 2x^{2} - x + 8x - 4 = 2x^{2} + 7x - 4$
3. Step 3
  A1 Collect and factorise the result to match the printed form.
  $2x^{2} + 9x + 10 - (2x^{2} + 7x - 4) = 2x^{2} + 9x + 10 - 2x^{2} - 7x + 4 = 2x + 14 = 2(x + 7)$
Answer
$(2x + 5)(x + 2) - (x + 4)(2x - 1) = 2x + 14 = 2(x + 7)$ , as required
Examiner tip
Examiner tip: "How much more" means A minus B. Write that as one expression first, with B's area in a bracket.

Common mistake: Changing only the first sign when subtracting: $2x^{2} + 9x + 10 - 2x^{2} + 7x - 4 = 16x + 6$ . The $-4$ must become $+4$ , or the "Show that" fails.
10Expanding two three-term brackets using a pattern [3 marks]
Stretch• Paper 1H/2H style — 3 marks• expanding brackets, difference of two squares
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Question
Expand and simplify $(x^{2} + 2x - 3)(x^{2} - 2x + 3)$ [3]
Step-by-step solution
1. Step 1
  M1 Notice that the brackets are $x^{2} + (2x - 3)$ and $x^{2} - (2x - 3)$ . So the product is a difference of two squares.
  $(x^{2} + (2x - 3))(x^{2} - (2x - 3)) = x^{4} - (2x - 3)^{2}$
2. Step 2
  M1 Expand the square as three terms.
  $(2x - 3)^{2} = 4x^{2} - 12x + 9$
3. Step 3
  A1 Subtract every term.
  $x^{4} - (4x^{2} - 12x + 9) = x^{4} - 4x^{2} + 12x - 9$
Answer
$x^{4} - 4x^{2} + 12x - 9$
Examiner tip
Examiner tip: The full nine-term expansion also earns all three marks, but spotting the pattern leaves far fewer places to go wrong. Check with $x = 1$ : $(0)(2) = 0$ and $1 - 4 + 12 - 9 = 0$ .

Common mistake: Writing $x^{4} - 4x^{2} - 12x + 9$ , which subtracts only the first term of $(2x - 3)^{2}$ . Brackets round the square prevent this.

Question 1
Paper 1H/2H style — 1 mark1 mark
Expand $-2y(3 - y)$ [1]
Model answer
$-2y \times 3 = -6y$ and $-2y \times (-y) = +2y^{2}$

Answer: $2y^{2} - 6y$
Why this scores
Mark scheme:
- B1 for $2y^{2} - 6y$ oe
Why this scores full marks: Both terms are multiplied by $-2y$ , including its sign. The usual wrong answer is $-6y - 2y^{2}$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
Expand and simplify $(4 - x)(2x - 3)$ [2]
Model answer
$(4 - x)(2x - 3) = 8x - 12 - 2x^{2} + 3x$

$= -2x^{2} + 11x - 12$
Why this scores
Mark scheme:
- M1 for 3 of the 4 terms $8x$ , $-12$ , $-2x^{2}$ , $+3x$ correct
- A1 for $-2x^{2} + 11x - 12$ oe
Why this scores full marks: All four products are written before collecting. The $-x$ carries its sign into both products: $-x \times 2x = -2x^{2}$ and $(-x) \times (-3) = +3x$ , not $-3x$ (which would give $5x$ ).
Question 3
Paper 1H/2H style — 2 marks2 marks
Simplify fully $(3x^{4}y)^{2} \times 2xy^{3}$ [2]
Model answer
$(3x^{4}y)^{2} = 9x^{8}y^{2}$

$9x^{8}y^{2} \times 2xy^{3} = 18x^{9}y^{5}$
Why this scores
Mark scheme:
- B2 for $18x^{9}y^{5}$ (B1 for $9x^{8}y^{2}$ seen, or for an answer of the form $kx^{m}y^{n}$ with two of $k = 18$ , $m = 9$ , $n = 5$ correct)
Why this scores full marks: The bracket is squared first, so the $3$ becomes $9$ and the power $4$ becomes $8$ . Then the numbers are multiplied and the powers added.
Question 4
Paper 1H/2H style — 2 marks2 marks
Expand and simplify $(5 - 3t)^{2}$ [2]
Model answer
$(5 - 3t)(5 - 3t) = 25 - 15t - 15t + 9t^{2}$

$= 25 - 30t + 9t^{2}$
Why this scores
Mark scheme:
- M1 for 3 of the 4 terms $25$ , $-15t$ , $-15t$ , $+9t^{2}$ correct
- A1 for $9t^{2} - 30t + 25$ oe
Why this scores full marks: Writing the bracket twice gives the two middle terms. $(-3t)^{2} = +9t^{2}$ , with the $3$ squared as well as the $t$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
Expand and simplify $4(2x - 1)^{2} - (3x + 2)(x - 5)$ [3]
Model answer
$(2x - 1)^{2} = 4x^{2} - 4x + 1$ , so $4(2x - 1)^{2} = 16x^{2} - 16x + 4$

$(3x + 2)(x - 5) = 3x^{2} - 15x + 2x - 10 = 3x^{2} - 13x - 10$

$16x^{2} - 16x + 4 - (3x^{2} - 13x - 10) = 16x^{2} - 16x + 4 - 3x^{2} + 13x + 10$

$= 13x^{2} - 3x + 14$
Why this scores
Mark scheme:
- M1 for $4x^{2} - 4x + 1$ or $3x^{2} - 13x - 10$
- M1 for $16x^{2} - 16x + 4$ and $3x^{2} - 13x - 10$ , with the second subtracted as a whole
- A1 for $13x^{2} - 3x + 14$ cao
Why this scores full marks: The square is expanded before multiplying by $4$ , rather than squaring $4(2x - 1) = 8x - 4$ . The second product is bracketed before it is subtracted, so $-10$ becomes $+10$ .
Question 6
Paper 1H/2H style — 3 marks3 marks
$(x + a)(x - 6) \equiv x^{2} + bx - 42$

Find the value of $a$ and the value of $b$ . [3]
Model answer
$(x + a)(x - 6) = x^{2} - 6x + ax - 6a = x^{2} + (a - 6)x - 6a$

Constants: $-6a = -42$ , so $a = 7$

$x$ terms: $b = a - 6 = 1$
Why this scores
Mark scheme:
- M1 for $x^{2} - 6x + ax - 6a$ oe, or $-6a = -42$
- A1 for $a = 7$
- A1 for $b = 1$ (ft their $a$ )
Why this scores full marks: The constant terms give $a$ straight away. The $x$ terms $-6x + ax$ are combined into $(a - 6)x$ before comparing with $bx$ . A check: $(x + 7)(x - 6) = x^{2} + x - 42$ .
Question 7
Paper 1H/2H style — 3 marks3 marks
Expand and simplify $(x - 5)^{2}(2x + 1)$ [3]
Model answer
$(x - 5)^{2} = (x - 5)(x - 5) = x^{2} - 5x - 5x + 25 = x^{2} - 10x + 25$

$(x^{2} - 10x + 25)(2x + 1) = 2x^{3} + x^{2} - 20x^{2} - 10x + 50x + 25$

$= 2x^{3} - 19x^{2} + 40x + 25$
Why this scores
Mark scheme:
- M1 for $(x - 5)^{2} = x^{2} - 10x + 25$ , or the product of any two of the three brackets correct
- M1 for multiplying by the remaining bracket, at least 4 of the 6 terms correct
- A1 for $2x^{3} - 19x^{2} + 40x + 25$ cao
Why this scores full marks: The square is treated as two brackets, $(x - 5)(x - 5)$ , so its middle term $-10x$ is not lost, and all six terms are written before collecting. Substituting $x = 1$ checks it: $(-4)^{2}(3) = 48$ and $2 - 19 + 40 + 25 = 48$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
A triangle has base $(4x + 2)$ cm and perpendicular height $(x + 3)$ cm.

A rectangle has length $(2x + 1)$ cm and width $(x + 1)$ cm.

Show that the area of the triangle is $2(2x + 1)$ cm² more than the area of the rectangle. [4]
Model answer
Triangle: $\dfrac{1}{2}(4x + 2)(x + 3) = (2x + 1)(x + 3) = 2x^{2} + 6x + x + 3 = 2x^{2} + 7x + 3$

Rectangle: $(2x + 1)(x + 1) = 2x^{2} + 2x + x + 1 = 2x^{2} + 3x + 1$

Difference: $2x^{2} + 7x + 3 - (2x^{2} + 3x + 1) = 4x + 2$

$= 2(2x + 1)$
Why this scores
Mark scheme:
- M1 for $\dfrac{1}{2}(4x + 2)(x + 3)$ oe expanded, e.g. $2x^{2} + 7x + 3$
- M1 for $2x^{2} + 3x + 1$
- M1 dep on the first M1, for subtracting the rectangle area from the triangle area
- A1 for a complete, correct argument leading to $2(2x + 1)$
Why this scores full marks: Halving $(4x + 2)$ first to get $(2x + 1)$ keeps the expansion simple. Every stage is written down, and the final line is put in the printed form.
Question 9
Paper 1H/2H style — 4 marks4 marks
A cuboid measures $x$ cm by $(x + 2)$ cm by $(x + 5)$ cm.

Show that the total surface area of the cuboid, in cm², is $6x^{2} + 28x + 20$ [4]
Model answer
The three different faces have areas

$x(x + 2) = x^{2} + 2x$

$x(x + 5) = x^{2} + 5x$

$(x + 2)(x + 5) = x^{2} + 7x + 10$

Sum: $3x^{2} + 14x + 10$

There are two of each face: $2(3x^{2} + 14x + 10) = 6x^{2} + 28x + 20$
Why this scores
Mark scheme:
- M1 for the area of any one face expanded correctly
- M1 for all three face areas correct
- M1 for $2 \times$ the sum of the three face areas oe
- A1 for a complete, correct argument leading to $6x^{2} + 28x + 20$
Why this scores full marks: The script lists the three different faces, then doubles their total, because opposite faces are equal. Weaker scripts find the volume, $x(x + 2)(x + 5)$ , instead of the surface area.
Question 10
Paper 1H/2H style — 5 marks5 marks
Show that $(x + 2)(x + 3)(x + 4) - x(x + 3)(x + 6) = 8(x + 3)$ [5]
Model answer
$(x + 2)(x + 4) = x^{2} + 6x + 8$

$(x^{2} + 6x + 8)(x + 3) = x^{3} + 3x^{2} + 6x^{2} + 18x + 8x + 24 = x^{3} + 9x^{2} + 26x + 24$

$x(x + 3)(x + 6) = x(x^{2} + 9x + 18) = x^{3} + 9x^{2} + 18x$

$x^{3} + 9x^{2} + 26x + 24 - (x^{3} + 9x^{2} + 18x) = 8x + 24$

$= 8(x + 3)$
Why this scores
Mark scheme:
- M1 for the product of any two brackets in the first term correct
- M1 for $x^{3} + 9x^{2} + 26x + 24$ (at least 3 of the 4 terms correct)
- M1 for $x^{3} + 9x^{2} + 18x$
- M1 dep on the previous M marks, for subtracting the whole second expansion, e.g. $8x + 24$
- A1 for a complete, correct argument leading to $8(x + 3)$
Why this scores full marks: Each triple product is expanded in two clear stages. The second is bracketed before it is subtracted, so the $x^{3}$ and $x^{2}$ terms cancel exactly, and the final line is factorised to match the printed form.

4MA1 Algebraic Manipulation

At a glance

What you’ll learn

Expanding brackets — FOIL and beyond

Simplifying — collecting like terms

Expression vs equation vs identity

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Expand a single bracket

2Expand a double bracket

3Square a binomial

4Expand three brackets

5Simplify by collecting like terms

6Writing an expression from a description [1 mark]

7Simplifying a quotient where a letter ends up in the denominator [2 marks]

8Multiplying a bracket by a three-term bracket [2 marks]

9How much larger one area is than another [3 marks]

10Expanding two three-term brackets using a pattern [3 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10