Mathematics · AQA 8300 §A2
Expanding and Factorising
- Expand single and double brackets correctly
- Recognise and apply perfect square and difference of two squares identities
- Factorise expressions including by HCF
- Factorise quadratics when $a=1$ and when $a\neq1$
- Verify factorisations by re-expanding
🔑 Core Concepts
Single Bracket Expansion
Definition
Expanding a single bracket means multiplying the term outside by every term inside. The distributive law states: $a(b+c) = ab + ac$.
Every term inside the bracket is multiplied by the factor outside. This includes the sign — a negative outside the bracket flips all signs inside.
Single Bracket Rule
$$a(b + c) = ab + ac$$
$a$ = multiplier
$b, c$ = terms inside
Common Mistake
$-3(2x - 5) \neq -6x - 15$. The minus outside applies to both terms: $-3 \times (-5) = +15$, so the answer is $-6x + 15$.
Exam Tip
Always check your sign on every term. Draw arrows to show which term multiplies which if you lose track.
Double Bracket Expansion (FOIL / Grid Method)
Definition
Expanding two brackets means multiplying each term in the first bracket by each term in the second. FOIL stands for First, Outer, Inner, Last.
Double Bracket General Form
$$(x + a)(x + b) = x^2 + (a+b)x + ab$$
$a, b$ = constants
Middle coefficient = $a+b$
Constant term = $a \times b$
The grid method is often safer for more complex expansions:
| Grid × | $x$ | $+b$ |
|---|---|---|
| $x$ | $x^2$ | $bx$ |
| $+a$ | $ax$ | $ab$ |
Exam Tip
After expanding, always collect like terms. The two middle terms ($bx$ and $ax$) combine to give $(a+b)x$.
Perfect Square Identities
A perfect square occurs when a bracket is squared. The result always has a specific structure — memorise it to avoid expanding fully each time and to spot factorisations instantly.
Perfect Square: Sum
$$(a+b)^2 = a^2 + 2ab + b^2$$
First² = $a^2$
Twice the product = $2ab$
Second² = $b^2$
Perfect Square: Difference
$$(a-b)^2 = a^2 - 2ab + b^2$$
Middle term is $-2ab$
Last term is always $+b^2$
Memory Trick
"Square–Double–Square": $(a \pm b)^2$ = (square the first) ± (double the product) + (square the last). The last term is ALWAYS positive.
Common Mistake
$(a+b)^2 \neq a^2 + b^2$. You must include the middle term $2ab$. This is one of the most common errors in GCSE algebra.
Difference of Two Squares
Definition
When two brackets are identical except for opposite signs, the inner and outer terms cancel, leaving the difference of two squares.
Difference of Two Squares
$$(a+b)(a-b) = a^2 - b^2$$
$a^2$ = first term squared
$b^2$ = second term squared
Middle terms cancel (+$ab$ and $-ab$)
Exam Tip
Use DOTS in reverse for factorising: any expression of the form $a^2 - b^2$ (no middle term, minus sign) factorises as $(a+b)(a-b)$. Watch for disguised forms like $x^2 - 16 = (x+4)(x-4)$.
Triple Bracket Expansion (Grade 9)
Definition
Expand three brackets by first expanding any two brackets, collecting like terms, then multiplying the resulting quadratic by the third bracket term by term.
Important
Always expand in stages. Never try to do all three at once. $(A)(B)(C)$ → expand $A \times B$ first → get quadratic → expand that by $C$ → collect all terms.
Exam Tip
The result of expanding three linear brackets is always a cubic expression $ax^3 + bx^2 + cx + d$. If you get a different degree, recheck your working.
Factorising by Highest Common Factor (HCF)
Definition
Factorising by HCF means finding the largest factor common to all terms and placing it outside a bracket. It is the reverse of single-bracket expansion.
Find HCF of all coefficients
→
Find lowest power of each variable present in every term
→
Write HCF outside bracket
→
Divide each term by HCF for inside
→
Check by re-expanding
Exam Tip
Always factorise fully — take out the highest common factor, not just any common factor. E.g. $6x^2 + 9x = 3x(2x+3)$ not $3(2x^2+3x)$.
Factorising Quadratics ($a = 1$)
Definition
To factorise $x^2 + bx + c$, find two numbers $p$ and $q$ such that $p \times q = c$ (constant) and $p + q = b$ (coefficient of $x$). Then write $(x+p)(x+q)$.
Factorising $x^2 + bx + c$
$$\text{Find } p, q \text{ where } p \times q = c \text{ and } p + q = b$$
Product = $c$
Sum = $b$
Result: $(x+p)(x+q)$
Common Mistake
When $c$ is negative, one of $p$, $q$ must be negative. Systematically list factor pairs of $c$ (with signs) until you find the pair that sums to $b$.
Factorising Quadratics ($a \neq 1$) — ac Method
Definition — ac Method
For $ax^2 + bx + c$ where $a \neq 1$: compute $ac$, find two numbers $p$ and $q$ where $p \times q = ac$ and $p + q = b$, split the $bx$ term into $px + qx$, then factorise by grouping.
ac Method Steps
$$ax^2 + bx + c \xrightarrow{p \cdot q = ac,\; p+q=b} ax^2 + px + qx + c$$
Step 1: compute $ac$
Step 2: find $p, q$
Step 3: split, group
Step 4: factorise groups
Multiply $a \times c$
→
Find $p \times q = ac$, $p + q = b$
→
Split: $ax^2 + px + qx + c$
→
Group pairs and factorise each
→
Extract common bracket
Exam Tip
After factorising, always verify by expanding your answer. On a 3-mark question you lose all marks for a wrong factorisation, so the 30-second check is always worth it.
Proving Identities by Expansion (Grade 9)
Definition
An identity (using the ≡ symbol) is true for all values of the variable. To prove an identity, start from one side (usually the more complex), expand and simplify, and show it equals the other side. Never "cross the equals sign."
Important — Grade 9
Algebraic proof questions use $\equiv$ to mean "identically equal." You must show full working from LHS to RHS (or vice versa) without assuming what you're trying to prove. Marks require correct algebraic manipulation, not just a numerical check.
Simplifying Algebraic Fractions by Factorising (Grade 9)
Definition
An algebraic fraction can be simplified by factorising the numerator and/or denominator, then cancelling any common factors. E.g. $\dfrac{x^2-9}{x+3} = \dfrac{(x+3)(x-3)}{x+3} = x-3$.
Common Mistake
You can only cancel factors (things multiplied), not terms (things added). $\dfrac{x^2 + 4}{x} \neq x + 4$, but $\dfrac{x(x+4)}{x} = x + 4$.
🗺️ Visual Notes
Expanding & Factorising
Expanding
- Single bracket: $a(b+c)=ab+ac$
- Double bracket: FOIL or grid
- Perfect squares: square–double–square
- Triple brackets: two stages
Identities
- $(a+b)^2 = a^2+2ab+b^2$
- $(a-b)^2 = a^2-2ab+b^2$
- $(a+b)(a-b) = a^2-b^2$
- Use ≡ for proofs
Factorising
- HCF: take out largest factor
- Quadratic $a=1$: find $p+q=b$, $pq=c$
- Quadratic $a\neq 1$: ac method
- DOTS: $a^2-b^2=(a+b)(a-b)$
Grade 9 Skills
- Triple bracket expansion
- Proving algebraic identities
- Simplifying algebraic fractions
- Non-integer coefficients
Checking
- Re-expand to verify factorisation
- Substitute a value to test identity
- Degree check (number of factors = degree)
- Sign check on constant term
Common Errors
- Missing $2ab$ in perfect squares
- Wrong sign when bracket is negative
- Partial HCF (not fully factorised)
- Cancelling terms not factors in fractions
Comparison: Factorising Methods
| Expression Form | Method | Example | Result |
|---|---|---|---|
| $ka + kb$ | HCF | $6x^2 + 9x$ | $3x(2x+3)$ |
| $x^2 + bx + c$ | Factor pairs ($a=1$) | $x^2+5x+6$ | $(x+2)(x+3)$ |
| $ax^2 + bx + c$ | ac method | $6x^2+7x+2$ | $(2x+1)(3x+2)$ |
| $a^2 - b^2$ | Difference of two squares | $x^2-25$ | $(x+5)(x-5)$ |
| $(a\pm b)^2$ | Perfect square recognition | $x^2+6x+9$ | $(x+3)^2$ |
| $a^2 \pm 2ab + b^2$ | Perfect square | $4x^2-12x+9$ | $(2x-3)^2$ |
Decision Tree: Which Factorising Method?
Is there a common factor in ALL terms?
Yes →
Take out HCF first, then continue
No →
Is it $a^2-b^2$?
Yes →
DOTS: $(a+b)(a-b)$
No →
Is $a=1$?
Yes →
Factor pair method
No →
ac method (split middle term)
✏️ Worked Examples
Grade 4–5 · Single & Double Brackets
(a) Expand and simplify $3(2x - 4) + 5(x + 1)$.
(b) Expand and simplify $(x + 3)(x - 7)$.
(b) Expand and simplify $(x + 3)(x - 7)$.
1
Part (a): Expand each bracket
$3(2x-4) = 6x - 12$ and $5(x+1) = 5x + 5$
2
Collect like terms
$6x - 12 + 5x + 5 = 11x - 7$
3
Part (b): FOIL method
F: $x \times x = x^2$ O: $x \times (-7) = -7x$ I: $3 \times x = 3x$ L: $3 \times (-7) = -21$
4
Collect like terms
$x^2 - 7x + 3x - 21 = x^2 - 4x - 21$
(a) $11x - 7$ (b) $x^2 - 4x - 21$
Grade 6–7 · Factorising Quadratics
(a) Factorise $x^2 - 3x - 18$.
(b) Factorise $6x^2 + 11x - 10$ using the ac method.
(b) Factorise $6x^2 + 11x - 10$ using the ac method.
1
Part (a): Find factor pairs of $-18$ that sum to $-3$
Pairs of $-18$: $(1,-18),(2,-9),(3,-6),(6,-3),(9,-2),\ldots$ We need sum $= -3$: $\mathbf{3 \times (-6) = -18}$ and $3 + (-6) = -3$. ✓
2
Write factorised form
$x^2 - 3x - 18 = (x + 3)(x - 6)$
3
Part (b): Compute $ac$
$a = 6,\ b = 11,\ c = -10.$ So $ac = 6 \times (-10) = -60$.
4
Find $p, q$ where $p \times q = -60$ and $p + q = 11$
Pairs of $-60$ summing to $11$: $\mathbf{15 \times (-4) = -60}$ and $15 + (-4) = 11$. ✓
5
Split and group
$6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5)$
(a) $(x+3)(x-6)$ (b) $(3x-2)(2x+5)$
Grade 9 · Triple Brackets & Algebraic Proof
(a) Expand and simplify $(x+2)(x-1)(x+3)$.
(b) Prove that $(2n+3)^2 - (2n-3)^2 \equiv 24n$ for all values of $n$.
(b) Prove that $(2n+3)^2 - (2n-3)^2 \equiv 24n$ for all values of $n$.
1
Part (a): Expand first two brackets
$(x+2)(x-1) = x^2 - x + 2x - 2 = x^2 + x - 2$
2
Multiply quadratic by third bracket
$(x^2+x-2)(x+3)$: $x^2 \cdot x = x^3$; $x^2 \cdot 3 = 3x^2$; $x \cdot x = x^2$; $x \cdot 3 = 3x$; $-2 \cdot x = -2x$; $-2 \cdot 3 = -6$
3
Collect like terms
$x^3 + 3x^2 + x^2 + 3x - 2x - 6 = x^3 + 4x^2 + x - 6$
4
Part (b): Expand LHS — start with $(2n+3)^2$
$(2n+3)^2 = 4n^2 + 12n + 9$
5
Expand $(2n-3)^2$
$(2n-3)^2 = 4n^2 - 12n + 9$
6
Subtract and simplify
$(4n^2+12n+9) - (4n^2-12n+9) = 4n^2+12n+9 - 4n^2+12n-9 = 24n$ ✓
(a) $x^3 + 4x^2 + x - 6$ (b) LHS $= (2n+3)^2 - (2n-3)^2 = (4n^2+12n+9)-(4n^2-12n+9) = 24n =$ RHS ✓ QED
❓ Exam Questions
Question 11 mark
Expand $-4(3x - 2)$.
Answer: $-12x + 8$
Mark scheme:
B1: $-12x + 8$ (both terms correct, accept $8 - 12x$)
Mark scheme:
B1: $-12x + 8$ (both terms correct, accept $8 - 12x$)
Question 22 marks
Expand and simplify $(2x + 5)(2x - 5)$.
Answer: $4x^2 - 25$
Mark scheme:
M1: Attempt to expand — at least 3 correct terms seen, e.g. $4x^2 - 10x + 10x - 25$
A1: $4x^2 - 25$ (cao, must be simplified — recognise DOTS)
Mark scheme:
M1: Attempt to expand — at least 3 correct terms seen, e.g. $4x^2 - 10x + 10x - 25$
A1: $4x^2 - 25$ (cao, must be simplified — recognise DOTS)
Question 33 marks
Factorise fully $12x^2y - 8xy^2$.
Answer: $4xy(3x - 2y)$
Mark scheme:
M1: Identifies $4xy$ as the HCF or shows partial factorisation e.g. $4(3x^2y - 2xy^2)$
M1: Correct structure of factorisation with $xy$ taken out
A1: $4xy(3x-2y)$ — both factors correct
Mark scheme:
M1: Identifies $4xy$ as the HCF or shows partial factorisation e.g. $4(3x^2y - 2xy^2)$
M1: Correct structure of factorisation with $xy$ taken out
A1: $4xy(3x-2y)$ — both factors correct
Question 43 marks
Factorise $3x^2 - 7x - 6$.
Answer: $(3x + 2)(x - 3)$
Mark scheme:
M1: Uses ac method — $ac = 3 \times (-6) = -18$; attempts to find $p+q=-7$, $pq=-18$
M1: Correctly identifies $p=2$, $q=-9$ and splits: $3x^2 + 2x - 9x - 6$
A1: $(3x+2)(x-3)$ — both brackets correct. Accept $(x-3)(3x+2)$.
Mark scheme:
M1: Uses ac method — $ac = 3 \times (-6) = -18$; attempts to find $p+q=-7$, $pq=-18$
M1: Correctly identifies $p=2$, $q=-9$ and splits: $3x^2 + 2x - 9x - 6$
A1: $(3x+2)(x-3)$ — both brackets correct. Accept $(x-3)(3x+2)$.
Question 54 marks
Simplify fully $\dfrac{x^2 - x - 12}{x^2 - 16}$.
Answer: $\dfrac{x+3}{x+4}$
Mark scheme:
M1: Factorise numerator — finds two numbers with product $-12$ and sum $-1$: $(-4)(3)$ → $(x-4)(x+3)$
M1: Factorise denominator using DOTS — $x^2-16=(x+4)(x-4)$
M1: Cancel common factor $(x-4)$
A1: $\dfrac{x+3}{x+4}$ (cao, fully simplified)
Mark scheme:
M1: Factorise numerator — finds two numbers with product $-12$ and sum $-1$: $(-4)(3)$ → $(x-4)(x+3)$
M1: Factorise denominator using DOTS — $x^2-16=(x+4)(x-4)$
M1: Cancel common factor $(x-4)$
A1: $\dfrac{x+3}{x+4}$ (cao, fully simplified)
Question 66 marks
(a) Expand and simplify $(x+1)(x-2)(x+4)$. [3]
(b) Hence, or otherwise, prove that $(x+1)(x-2)(x+4) + 8 \equiv x(x^2 + 3x - 2)$ is false, and find the correct constant $k$ such that $(x+1)(x-2)(x+4) + k \equiv x(x^2+3x-2)$. [3]
Part (a):
$(x+1)(x-2) = x^2 - x - 2$
$(x^2-x-2)(x+4) = x^3+4x^2-x^2-4x-2x-8 = x^3+3x^2-6x-8$
Part (b):
LHS with $k=8$: $x^3+3x^2-6x-8+8 = x^3+3x^2-6x$
RHS: $x(x^2+3x-2) = x^3+3x^2-2x$
These are not equal ($-6x \neq -2x$), so $k=8$ is incorrect.
For equality: $x^3+3x^2-6x+k = x^3+3x^2-2x$ → $-6x+k = -2x$ → $k = 4x$ — but $k$ must be a constant, so no constant $k$ makes them identically equal. (Award marks for showing the contradiction clearly.)
Mark scheme:
M1 A1 A1 for part (a); M1 for comparing expansions, M1 for identifying discrepancy, A1 for conclusion with justification.
$(x+1)(x-2) = x^2 - x - 2$
$(x^2-x-2)(x+4) = x^3+4x^2-x^2-4x-2x-8 = x^3+3x^2-6x-8$
Part (b):
LHS with $k=8$: $x^3+3x^2-6x-8+8 = x^3+3x^2-6x$
RHS: $x(x^2+3x-2) = x^3+3x^2-2x$
These are not equal ($-6x \neq -2x$), so $k=8$ is incorrect.
For equality: $x^3+3x^2-6x+k = x^3+3x^2-2x$ → $-6x+k = -2x$ → $k = 4x$ — but $k$ must be a constant, so no constant $k$ makes them identically equal. (Award marks for showing the contradiction clearly.)
Mark scheme:
M1 A1 A1 for part (a); M1 for comparing expansions, M1 for identifying discrepancy, A1 for conclusion with justification.
⭐ Grade 9 Model Answers
Focus Question — Algebraic Proof (Grade 9)
Prove that the difference of the squares of any two consecutive odd numbers is always a multiple of 8.
Grade 9 · Full Proof
Prove that the difference of the squares of any two consecutive odd numbers is divisible by 8.
1
Set up algebra — define consecutive odd numbers
Let the smaller odd number be $2n+1$ and the next consecutive odd number be $2n+3$ (where $n$ is an integer).
2
Square each expression using perfect square identity
$(2n+3)^2 = 4n^2 + 12n + 9$
$(2n+1)^2 = 4n^2 + 4n + 1$
$(2n+1)^2 = 4n^2 + 4n + 1$
3
Find the difference
$(2n+3)^2 - (2n+1)^2 = (4n^2+12n+9) - (4n^2+4n+1)$
$= 4n^2 + 12n + 9 - 4n^2 - 4n - 1$
$= 8n + 8$
$= 4n^2 + 12n + 9 - 4n^2 - 4n - 1$
$= 8n + 8$
4
Factorise and conclude
$8n + 8 = 8(n+1)$
Since $n$ is an integer, $(n+1)$ is also an integer. Therefore $8(n+1)$ is always a multiple of 8. ✓
Since $n$ is an integer, $(n+1)$ is also an integer. Therefore $8(n+1)$ is always a multiple of 8. ✓
QED: The difference of the squares of any two consecutive odd numbers $= 8(n+1)$, which is divisible by 8 for all integer values of $n$.
Why Each Part Earns Marks
| Step | What Was Done | Marks Earned |
|---|---|---|
| 1 | General consecutive odd numbers defined as $2n+1$ and $2n+3$ | M1 — correct algebraic representation |
| 2 | Both squares correctly expanded using $(a+b)^2 = a^2+2ab+b^2$ | M1 — correct expansions |
| 3 | Subtraction carried out with correct signs throughout | M1 — correct difference |
| 4 | Result written as $8(n+1)$ and conclusion stated explicitly | A1 — fully factoried with conclusion |
Grade 9 Proof Technique
Always end a proof with an explicit conclusion. Write "Therefore…" and restate what you have shown. Examiners need to see the logical conclusion, not just the algebra.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Expand | Remove brackets by multiplying |
| Factorise | Write as a product of factors |
| HCF | Highest common factor |
| DOTS | Difference of two squares |
| Identity (≡) | True for all values |
| ac method | Split middle term for $a \neq 1$ |
Essential Formulae
- $a(b+c) = ab+ac$
- $(a+b)^2 = a^2+2ab+b^2$
- $(a-b)^2 = a^2-2ab+b^2$
- $(a+b)(a-b) = a^2-b^2$
- $(x+p)(x+q) = x^2+(p+q)x+pq$
- ac: $pq=ac$, $p+q=b$
Memory Hooks
- FOIL: First, Outer, Inner, Last
- Square–Double–Square for $(a\pm b)^2$
- DOTS: no middle term = difference of squares
- Signs in DOTS: one plus, one minus
- ac: always multiply $a \times c$, not $a \times b$
- Triple bracket: always two stages
Exam Tips
- Always re-expand to check your factorisation
- Factorise fully — take out the HCF first
- $(a+b)^2 \neq a^2+b^2$ — include $2ab$
- In proofs, never assume what you're proving
- Cancel factors not terms in fractions
- Final term of perfect square is always $+b^2$
🔄 Flashcards
Click a card to reveal the answer.
✗ Common Mistakes
Mistake 1: Forgetting the middle term in a perfect square
Wrong: $(x+5)^2 = x^2 + 25$Why marks lost: The $2ab$ term is missing — this is a fundamental error that loses all marks on an identity question.
Correct: $(x+5)^2 = x^2 + 10x + 25$. Use the rule: square first, double the product, square last.
Mistake 2: Wrong sign when expanding a negative bracket
Wrong: $-2(x-3) = -2x - 6$Why marks lost: $-2 \times (-3)$ is $+6$, not $-6$. Sign errors cause knock-on mistakes in subsequent working.
Correct: $-2(x-3) = -2x + 6$. Always multiply the sign as well as the coefficient.
Mistake 3: Incomplete factorisation (partial HCF)
Wrong: $12x^2 + 8x = 4(3x^2 + 2x)$ — not fully factorisedWhy marks lost: The instruction "factorise fully" requires the HCF, not just a common factor. This scores 1 out of 2 at best.
Correct: $12x^2 + 8x = 4x(3x + 2)$ — both the numerical HCF and the variable factor $x$ are extracted.
Mistake 4: Cancelling terms (not factors) in algebraic fractions
Wrong: $\dfrac{x^2 + 6x}{x} = x + 6x = 7x$Why marks lost: You can only cancel common factors — things multiplied. Here, you must first write $\dfrac{x(x+6)}{x} = x+6$.
Correct: Factorise the numerator first, then cancel the factor.
Mistake 5: Using $ac$ as $a \times b$ instead of $a \times c$ in the ac method
Wrong (for $6x^2+11x-10$): Computing $a \times b = 6 \times 11 = 66$ instead of $ac = 6 \times (-10) = -60$Why marks lost: This gives completely wrong factor pairs, leading to an incorrect factorisation.
Correct: In $ax^2 + bx + c$, multiply the first and last coefficients: $a \times c$.
Mistake 6: Incorrect second expansion in triple brackets
Wrong: Multiplying only the first term of the quadratic by the third bracket — missing termsWhy marks lost: Every term of the quadratic must multiply every term of the linear factor — a systematic approach (grid or column) prevents missing terms.
Correct: Set out the multiplication clearly: each of the 3 terms of the quadratic by each of the 2 terms of the bracket = 6 individual products.
✅ Final Checklist
Click each item to mark it complete.
- I can expand a single bracket, including with a negative multiplier
- I can expand double brackets using FOIL or the grid method
- I can apply $(a+b)^2 = a^2+2ab+b^2$ without expanding from scratch
- I can apply $(a-b)^2 = a^2-2ab+b^2$ and know the last term is always $+b^2$
- I can recognise and apply the difference of two squares $(a+b)(a-b)=a^2-b^2$
- I can expand three brackets in two stages to get a cubic
- I can factorise by taking out the highest common factor fully
- I can factorise $x^2+bx+c$ by finding a factor pair with correct product and sum
- I can factorise $ax^2+bx+c$ ($a \neq 1$) using the ac method
- I can recognise and factorise a difference of two squares expression
- I can verify any factorisation by re-expanding
- I can simplify an algebraic fraction by factorising numerator and denominator
- I can construct an algebraic proof using expanding and factorising
- I know the difference between an equation (=) and an identity (≡)
- I can spot and avoid the six most common errors in this topic