VCE Vault

Year 9 Mathematics · Number and Algebra · 18m read

Algebraic Expansion and Factorisation

Ever noticed how a folded piece of paper and an unfolded one are the same thing, just in different forms? Expansion and factorisation work the same way—one multiplies out brackets to reveal what's inside, while the other finds what's common and bundles it back up. You'll learn to expand binomials (like (x + 3)(x + 2)) and reverse the process by factorising quadratics, skills that unlock harder algebra and are essential for solving equations in VCE maths.

Learning objective: Expand binomials and factorise expressions including monic quadratics.

By the end of this lesson you can:

  • I expand using distributive law and FOIL.
  • I factorise using common factors.
  • I factorise simple quadratics.

Expanding unpacks; factorising repacks. Expansion takes everything out of the brackets-box. Factorising finds the common pieces and packs them neatly back into brackets.

Expand and factor

EXPANDING — opening brackets.

### Single Bracket (Distributive Law)
$a(b + c) = ab + ac$

Example: $3(x + 4) = 3x + 12$

### Two Binomials (FOIL)
$(a + b)(c + d) = ac + ad + bc + bd$

FOIL stands for: First, Outer, Inner, Last terms.
$(x + 3)(x + 5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15$

### Special Products — memorise these!

| Identity | Expanded form |
|----------|--------------|
| $(a + b)^2$ | $a^2 + 2ab + b^2$ |
| $(a - b)^2$ | $a^2 - 2ab + b^2$ |
| $(a + b)(a - b)$ | $a^2 - b^2$ (DOTS) |

Examples:
• $(x + 4)^2 = x^2 + 8x + 16$
• $(2x - 3)^2 = 4x^2 - 12x + 9$
• $(x + 7)(x - 7) = x^2 - 49$

### FACTORISING — folding back into brackets.

Step 1: Look for a COMMON FACTOR first.
$6x^2 + 9x = 3x(2x + 3)$

Step 2: Difference of Two Squares (DOTS).
$a^2 - b^2 = (a + b)(a - b)$
$x^2 - 16 = (x + 4)(x - 4)$
$9x^2 - 25 = (3x + 5)(3x - 5)$

Step 3: Monic quadratics $x^2 + bx + c$.
Find two numbers $p$ and $q$ such that:
• $p \times q = c$ (product equals constant)
• $p + q = b$ (sum equals middle coefficient)
Then: $x^2 + bx + c = (x + p)(x + q)$

Example: $x^2 + 5x + 6$ → need product 6, sum 5 → use 2 and 3.
$x^2 + 5x + 6 = (x + 2)(x + 3)$

Example: $x^2 - 7x + 12$ → need product 12, sum −7 → use −3 and −4.
$x^2 - 7x + 12 = (x - 3)(x - 4)$

Example (negative c): $x^2 + 2x - 15$ → product −15, sum 2 → use +5 and −3.
$x^2 + 2x - 15 = (x + 5)(x - 3)$

### Verification
ALWAYS expand your factorised answer to check you get the original expression back.

### Australian context
A rectangular block of land has area $x^2 + 11x + 30$ m². The dimensions must be $(x + 5)$ m and $(x + 6)$ m — factorisation finds the dimensions!

Expansion Identities

  • Difference of two squares: a^2 - b^2 = (a+b)(a-b)
  • Perfect square (sum): (a+b)^2 = a^2 + 2ab + b^2
  • Perfect square (difference): (a-b)^2 = a^2 - 2ab + b^2
  • Monic quadratic: x^2 + bx + c = (x+p)(x+q) (where $pq=c$ and $p+q=b$)

Quick reminders

  • Signs matter: Watch +/− carefully. $(x-3)(x-4) \neq (x+3)(x+4)$.
  • HCF first: ALWAYS factor out the highest common factor before anything else.
  • DOTS check: Is it $a^2 - b^2$? If so, $(a+b)(a-b)$ immediately.

Worked example (Medium)

Factorise $x^2 - 7x + 12$.

  1. Identify. Monic quadratic: need two numbers with product $+12$ and sum $-7$.
  2. Trial. $(-3) \times (-4) = 12$ ✓ and $(-3) + (-4) = -7$ ✓
  3. Write. $(x - 3)(x - 4)$
  4. Verify. Expand: $x^2 - 4x - 3x + 12 = x^2 - 7x + 12$ ✓

Answer: $(x - 3)(x - 4)$

Worked example (Hard)

Factorise fully: $3x^3 - 12x$.

  1. HCF. Common factor is $3x$: $3x^3 - 12x = 3x(x^2 - 4)$
  2. DOTS. $x^2 - 4 = x^2 - 2^2 = (x+2)(x-2)$
  3. Final. $3x(x+2)(x-2)$
  4. Check. Expand $3x(x+2)(x-2) = 3x(x^2 - 4) = 3x^3 - 12x$ ✓

Answer: $3x(x+2)(x-2)$

Common mistakes

  • Expanding $(x+3)^2 = x^2 + 9$ (forgetting the middle term) $(x+3)^2 = (x+3)(x+3)$, which by FOIL gives THREE terms, not two. The middle term $2 \times x \times 3 = 6x$ is commonly dropped. Fix: Use the identity: $(a+b)^2 = a^2 + 2ab + b^2$, so $(x+3)^2 = x^2 + 6x + 9$.
  • Factorising $x^2 + 5x + 6$ as $(x+6)(x+1)$ — not checking both conditions $(x+6)(x+1)$: product = 6 ✓ but sum = 7 ✗. Both conditions must hold. Fix: Always verify: product of constants equals $c$, sum equals $b$. Here: $2 \times 3 = 6$ and $2 + 3 = 5$, so $(x+2)(x+3)$ is correct.
  • Writing $a^2 - b^2 = (a-b)^2$ $(a-b)^2 = a^2 - 2ab + b^2$, which has a middle term. The difference of two squares factors into TWO DIFFERENT binomials. Fix: $a^2 - b^2 = (a+b)(a-b)$. Verify by expanding: $(a+b)(a-b) = a^2 - ab + ab - b^2 = a^2 - b^2$ ✓

Practice questions

  1. Expand $(x + 3)(x + 5)$.

    1. $x^2 + 15$
    2. $x^2 + 8x + 15$
    3. $x^2 + 8x + 8$
    4. $2x + 15$

    (1 marks)

  2. Factorise $x^2 - 25$.

    (2 marks)

  3. Expand $(2x - 3)^2$.

    (2 marks)

  4. Factorise $x^2 + 3x - 10$.

    (2 marks)

  5. Factorise fully $2x^2 - 18$.

    (3 marks)

Going Deeper — Completing the Square

Completing the square is a technique that rewrites $x^2 + bx + c$ in the form $(x + h)^2 + k$.

Method:
1. Start with $x^2 + bx + c$.
2. Add and subtract $(b/2)^2$:
$x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$
3. The first three terms form a perfect square:
$\left(x + \frac{b}{2}\right)^2 + c - \left(\frac{b}{2}\right)^2$

Example: $x^2 + 6x + 5$
$= x^2 + 6x + 9 - 9 + 5$
$= (x+3)^2 - 4$

This form immediately tells us the vertex of the parabola $y = (x+3)^2 - 4$ is at $(-3, -4)$ — a key Year 10 technique previewed here.

Summary

  • FOIL expands two binomials: $(a+b)(c+d) = ac + ad + bc + bd$.
  • Three special identities: DOTS, and two perfect squares — memorise them.
  • To factorise: HCF first, then DOTS, then monic quadratic factor-finding.
  • Monic quadratic: find $p, q$ where $pq = c$ and $p + q = b$.
  • ALWAYS verify by expanding back to the original.

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