Overview
What you’ll learn
- Expand and simplify products of two linear brackets, collecting like terms.
- Use the identities (a + b)², (a − b)² and (a + b)(a − b) to expand quickly.
- Factorise expressions by taking out a common factor and by grouping.
- Factorise x² + px + q and the difference of two squares.
- Solve quadratic equations of the form x² + px + q = 0 by factorisation.
- Read quadratic graphs: parabola shape, x-intercepts (roots) and y-intercept.
In 30 seconds
- 1Expand every term into every term: (x + 2)(x + 5) = x² + 5x + 2x + 10 = x² + 7x + 10.
- 2The square identities: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b² — the middle term is twice the product.
- 3Difference of two squares: a² − b² = (a + b)(a − b). No middle term is the giveaway.
- 4To factorise x² + px + q, hunt two numbers that multiply to q and add to p.
- 5A product equals zero only when a factor equals zero: if (x − 3)(x + 5) = 0, then x = 3 or x = −5.
- 6The graph of a quadratic is a parabola; its x-intercepts are exactly the solutions of the equation.
Notes · 1 of 7
Expanding two brackets
Every term meets every term; then like terms tidy up.
A quadratic expression contains a square term like x² as its highest power. Multiplying two linear brackets produces one: in (x + 2)(x + 5), each term of the first bracket multiplies each term of the second — x·x, x·5, 2·x and 2·5 — giving x² + 5x + 2x + 10, which simplifies to x² + 7x + 10.
| Product | Result |
|---|---|
| x × x | x² |
| x × 5 | 5x |
| 2 × x | 2x |
| 2 × 5 | 10 |
The two middle terms are where marks are lost with negatives. In (x − 2)(x + 6): x² + 6x − 2x − 12 = x² + 4x − 12. The sign travels with its term — −2 times 6 is −12, and 6x plus −2x leaves +4x.
Notes · 2 of 7
Three identities worth memorising
Squared brackets and matched brackets have shortcuts — with trap checks built in.
| Identity | Expansion | Memory hook |
|---|---|---|
| (a + b)² | a² + 2ab + b² | square both, plus twice the product |
| (a − b)² | a² − 2ab + b² | same shape, middle term negative |
| (a + b)(a − b) | a² − b² | the middles cancel: difference of two squares |
The classic error is claiming (x + 4)² = x² + 16 — the middle term 2ab does not vanish. Fully expanded, (x + 4)² = x² + 8x + 16. The identities are shortcuts, not new rules: expanding (x + 4)(x + 4) the long way reaches the same answer.
Notes · 3 of 7
Factorising: common factors and grouping
Pull out what every term shares; split four terms into two friendly pairs.
Factorising reverses expansion. The first move is always the common factor: in 2x² + 6x both terms share 2x, so 2x² + 6x = 2x(x + 3). Check by expanding back — 2x times x and 2x times 3 rebuild the original.
With four terms, try grouping: pair them so each pair exposes a shared bracket. In ax + ay + bx + by, the first pair gives a(x + y) and the second b(x + y), leaving a(x + y) + b(x + y) = (a + b)(x + y). The shared bracket becomes the second factor.
Notes · 4 of 7
Factorising x² + px + q
Two numbers: they multiply to the last term and add to the middle coefficient.
- 11. Read p and q
In x² + 7x + 12, p = 7 and q = 12.
- 22. List factor pairs of q
12: 1×12, 2×6, 3×4 — and the negative versions if q is negative.
- 33. Pick the pair adding to p
3 + 4 = 7, so the pair is 3 and 4.
- 44. Write the brackets
(x + 3)(x + 4). Expand back to confirm.
| q | p | Both numbers are… | Example |
|---|---|---|---|
| positive | positive | positive | x² + 7x + 12 → (x + 3)(x + 4) |
| positive | negative | negative | x² − 9x + 20 → (x − 4)(x − 5) |
| negative | either | opposite signs | x² + 2x − 24 → (x + 6)(x − 4) |
Notes · 5 of 7
Recognising the difference of two squares
Two squares, minus sign, no middle term: it always splits into a matched pair.
An expression of the form a² − b² factorises instantly: a² − b² = (a + b)(a − b). So x² − 49 = (x + 7)(x − 7), and 4x² − 25 = (2x + 5)(2x − 5), since 4x² = (2x)². The tell-tale sign is a subtraction with nothing in the middle.
The reverse direction matters in expansions too: meeting (x + 9)(x − 9), the middle terms cancel and the answer is x² − 81 — no collecting needed. Knowing the identity in both directions saves time in both topics.
Notes · 6 of 7
Solving quadratic equations by factorisation
Get zero on one side, factorise, then let each factor take its turn at zero.
The engine of this chapter is the zero-product idea: if two things multiply to zero, at least one of them is zero. So (x − 3)(x + 5) = 0 forces x − 3 = 0 or x + 5 = 0, giving x = 3 or x = −5. Quadratics usually have two solutions; linear equations have one.
- 11. Make zero
Rearrange so one side is exactly 0 — never factorise an equation that does not equal 0.
- 22. Factorise
Use any method from this chapter: common factor, x² + px + q, or difference of squares.
- 33. Zero each factor
Set each bracket to 0 and solve the two mini-equations.
- 44. State both answers
Write the pair with “or”; check one root by substitution if time allows.
Notes · 7 of 7
Quadratic graphs: the parabola
A squared term bends the line into a U — and the roots sit on the x-axis.
| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| y | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
Plotting these points traces a symmetric U-shape called a parabola. It crosses the x-axis where y = 0 — for y = x² − 4 that is x = −2 and x = 2, exactly the solutions of x² − 4 = 0. The y-intercept comes from x = 0: here, −4.
In factored form y = (x − 2)(x − 8), the graph meets the x-axis at x = 2 and x = 8 — the roots jump out of the brackets. The curve turns around halfway between the roots, which is also where its lowest (or highest) point sits.
Revise
Key terms
The words examiners expect you to use precisely.
- Quadratic expression
- An expression whose highest power is a square term, such as x² + 7x + 10.
- Expand
- Multiply out brackets so no brackets remain.
- Partial products
- The four individual multiplications produced by expanding two linear brackets.
- Identity
- An equality true for every value of the letters, such as (a + b)² = a² + 2ab + b².
- Factorise
- Write an expression as a product of factors; the reverse of expanding.
- Common factor
- A term that divides every term of an expression, taken out as one factor.
Show 6 more key termsShow fewer key terms
- Grouping
- A factorising method for four terms: pair them, factor each pair, then factor the shared bracket.
- Difference of two squares
- An expression a² − b², which factors as (a + b)(a − b).
- Root
- A value of x that makes an equation equal zero; on a graph, an x-intercept.
- Zero-product idea
- If a product equals zero, at least one factor must equal zero.
- Parabola
- The U-shaped graph of a quadratic function; it turns around at its minimum or maximum point.
- Turning point
- The point where a parabola changes direction, halfway between its roots.
Revise
Common mistakes
Where students most often lose marks in this chapter — and what to do instead.
Mistake: Expanding (x + 4)² as x² + 16.
The middle term 2ab is skipped.
Instead: (x + 4)² = x² + 8x + 16 — square the first, add twice the product, square the last.
Mistake: Losing a negative when expanding (x − 2)(x + 6).
The −2 must multiply both x and 6.
Instead: x² + 6x − 2x − 12 = x² + 4x − 12; write all four partial products first.
Show 8 more mistakesShow fewer mistakes
Mistake: Claiming (x + 7)(x − 7) = x² + 49.
The middle terms cancel, and the last product is negative times positive.
Instead: Difference of two squares: (a + b)(a − b) = a² − b², so the answer is x² − 49.
Mistake: Choosing factor pairs that multiply correctly but add wrongly, e.g. (x + 2)(x + 6) for x² + 8x + 12.
The pair must satisfy both conditions: multiply to q AND add to p.
Instead: For x² + 8x + 12 the pair is 2 and 6 — but for x² + 7x + 12 it is 3 and 4. Always re-add.
Mistake: Trying to factorise an equation that does not equal zero.
The zero-product idea only works when the other side is exactly 0.
Instead: Rearrange first: x² + 2x − 24 = 12 must become x² + 2x − 36 = 0 before any factorising.
Mistake: Dropping one of the two solutions, e.g. answering only x = 4 for (x − 4)(x + 9) = 0.
Either factor can be the zero, so a quadratic generally has two roots.
Instead: Set every factor to zero and state the pair: x = 4 or x = −9.
Mistake: Reporting a root that makes a length or width negative.
The algebra is valid but the context is not — measurements cannot be negative.
Instead: Solve both roots, then reject the one that breaks the story and say why.
Mistake: Setting each bracket of (x − 2)(x + 5) = 7 to 7, e.g. x − 2 = 7 and x + 5 = 7.
The zero-product idea needs zero on the other side — a product equals 7 in many ways, none of them “each factor is 7”.
Instead: Expand, move the 7 across to get = 0, then factorise the new quadratic.
Mistake: Rewriting x² − 16 as (x − 4)².
A square expands with a middle term: (x − 4)² = x² − 8x + 16, which is not x² − 16.
Instead: x² − 16 is a difference of two squares: (x + 4)(x − 4).
Mistake: Insisting x² − 6x + 9 = 0 has two different solutions.
The left side is (x − 3)², and a square equals zero only when the bracket itself is zero — once.
Instead: (x − 3)² = 0 gives the single repeated root x = 3.
Check yourself
Quick quiz
18 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
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- Question 3: not answered
Expand two brackets
Expand (x + 2)(x + 5).
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Practise
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Exam-style practice in three tiers, from recall to data questions.
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Each school sets its own scope — check your assessment outline.
- Term 1Coming soonWeighted Assessment 1 (WA1)
- Term 4Coming soonEnd-of-Year Examination (EOY)
Questions
Frequently asked questions
What is Sec 2 Math Chapter 3 about?
Chapter 3 covers quadratic expressions and equations: expanding two brackets, the square and difference-of-squares identities, factorising by common factor, grouping and the two-number method, solving quadratic equations by factorisation, and a first look at parabolas.
How do I factorise x² + px + q?
Find two numbers that multiply to q and add to p, then write (x + first)(x + second). For x² + 7x + 12, the numbers 3 and 4 work: (x + 3)(x + 4). When q is negative, the two numbers have opposite signs.
Why does a quadratic equation usually have two solutions?
Because a product equals zero when either factor is zero. In (x − 3)(x + 5) = 0, either bracket can be the zero, giving x = 3 or x = −5. On a graph these are the two places the parabola crosses the x-axis.
How do I expand (x + 4)² correctly?
Use the square identity: (a + b)² = a² + 2ab + b², so (x + 4)² = x² + 8x + 16. The common wrong answer x² + 16 drops the middle term 8x — remember the middle term is twice the product.
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Chapter 4: Algebraic Fractions and Formulae