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Lesson 12 – The Distributive Property

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LESSON 12 / 100  |  12% COMPLETE  |  STAGE I — FOUNDATIONS
Stage I — Foundations

The Distributive Property

Expanding brackets — the mechanism that unlocks every equation to come.


The Core Law

The distributive property states that a factor multiplying a sum (or difference) inside brackets is applied to every term inside. The bracket is expanded; the factor is distributed.

a(b + c) = ab + ac
a(b − c) = ab − ac
3(x + 5)
↙        ↘
3×x + 3×5
= 3x + 15
Distributing is like handing out identical instructions to a group: if a coach tells each player to do 3 sets of push-ups and 3 sets of sprints, it is the same as saying 3 × (push-ups + sprints). Every member of the group gets the full instruction.
Key Point — Every Term Gets Multiplied

The factor outside the brackets multiplies every single term inside — not just the first. This is the most common place for errors: students multiply the first term and forget the rest.

✗ Wrong:   4(x + 3) = 4x + 3

✓ Correct:   4(x + 3) = 4x + 12


Expanding with Positive Factors

Worked Example A — Expand 5(2x − 3)
1
Multiply 5 by each term: 5 × 2x and 5 × (−3)
2
10x and −15
3
Result: 10x − 15
Worked Example B — Expand 3x(4x² − 2x + 1)
1
3x × 4x² = 12x³
2
3x × (−2x) = −6x²
3
3x × 1 = 3x
4
Result: 12x³ − 6x² + 3x
Multiplying Powers — Quick Reminder

When multiplying terms with the same base, add the exponents:

x × x² = x^(1+2) = x³

3x × 4x² = (3×4)(x^1 × x^2) = 12x³


Expanding with Negative Factors

When the factor outside the brackets is negative, every sign inside flips. This is where most errors occur.

−a(b + c) = −ab − ac
−a(b − c) = −ab + ac
Worked Example C — Expand −4(3x − 7)
1
−4 × 3x = −12x
2
−4 × (−7) = +28
3
Result: −12x + 28
Distributing a Bare Negative Sign

When only a minus sign precedes the brackets, the factor is −1. Every term inside changes sign:

−(x − 5) = −1(x − 5) = −x + 5

Students who drop the negative and write x − 5 unchanged make the single most common expansion error in all of algebra.


Expand and Simplify

The real power of the distributive property emerges when you combine it with collecting like terms. The standard task: expand all brackets, then collect like terms.

Worked Example D — Expand and simplify 3(2x + 1) + 4(x − 5)
1
Expand first bracket: 3(2x + 1) = 6x + 3
2
Expand second bracket: 4(x − 5) = 4x − 20
3
Combine: 6x + 3 + 4x − 20
4
Collect: (6x + 4x) + (3 − 20)
5
Result: 10x − 17
Worked Example E — Expand and simplify 5(a − 2b) − 2(3a + b)
1
5(a − 2b) = 5a − 10b
2
−2(3a + b) = −6a − 2b
3
Combine: 5a − 10b − 6a − 2b
4
a-terms: 5a − 6a = −a.   b-terms: −10b − 2b = −12b
5
Result: −a − 12b

Factorising — The Reverse Process

Factorising is the inverse of expanding. Instead of removing brackets, you introduce them by identifying a common factor across all terms and pulling it out.

ab + ac = a(b + c)
Worked Example F — Factorise 12x² − 8x
1
Find the GCF of 12 and 8: GCF = 4. Common variable: x. So the HCF of both terms is 4x.
2
Divide each term: 12x² ÷ 4x = 3x   and   8x ÷ 4x = 2
3
Write as product: 4x(3x − 2)
4
Verify by expanding: 4x(3x − 2) = 12x² − 8x
Verification Rule

Always verify a factorisation by re-expanding. If expansion returns the original expression, the factorisation is correct. This habit catches errors instantly.


The Distributive Property in Arithmetic

The distributive property also explains a mental arithmetic technique you may already use intuitively:

Mental Arithmetic — Calculate 7 × 98
1
Rewrite: 7 × (100 − 2)
2
Distribute: 7 × 100 − 7 × 2 = 700 − 14
3
Result: 686
Every time you mentally split a difficult multiplication into two easier ones, you are applying the distributive property. Algebra simply names and formalises a calculation technique the mind already uses.

Practice Set

1. Expand: 6(3x − 4)
6 × 3x = 18x.   6 × (−4) = −24.   Answer: 18x − 24
2. Expand: −3(5 − 2x)
−3 × 5 = −15.   −3 × (−2x) = +6x.   Answer: −15 + 6x (or 6x − 15)
3. Expand: 2x(x² − 5x + 3)
2x × x² = 2x³.   2x × (−5x) = −10x².   2x × 3 = 6x.   Answer: 2x³ − 10x² + 6x
4. Expand and simplify: 4(x + 3) − 2(x − 1)
4x + 12 − 2x + 2.   Collect: (4x − 2x) + (12 + 2) = 2x + 14
5. Expand and simplify: 3(2a − b) − (a + 4b)
6a − 3b − a − 4b.   a-terms: 5a.   b-terms: −7b.   Answer: 5a − 7b
6. Factorise fully: 15x³ − 10x²
HCF = 5x².   15x³ ÷ 5x² = 3x.   10x² ÷ 5x² = 2.   Answer: 5x²(3x − 2).   Verify: 5x²(3x − 2) = 15x³ − 10x² ✓
7. A square room has side length (x + 4) metres. A rectangular alcove of dimensions 2 × (x − 1) is removed from one corner. Write a simplified expression for the remaining floor area.
Square area = (x + 4)² — note: this requires lesson 13 (substitution) to fully expand. For now: Square area = (x+4)(x+4).   Alcove area = 2(x − 1) = 2x − 2.   Remaining = (x+4)(x+4) − (2x − 2).   At this stage, accept: (x+4)² − 2(x−1) as the simplified expression, noting (x+4)² will be expanded in a later lesson.
8. Challenge: Expand and simplify: 2x(x − 3) − x(x + 4) + 3(x² − 2)
2x(x − 3) = 2x² − 6x.   −x(x + 4) = −x² − 4x.   3(x² − 2) = 3x² − 6.   Combine: 2x² − 6x − x² − 4x + 3x² − 6.   x²-terms: 2 − 1 + 3 = 4x².   x-terms: −6x − 4x = −10x.   constants: −6.   Answer: 4x² − 10x − 6

Lesson Checklist

You Can Now
  • State and apply the distributive property: a(b + c) = ab + ac
  • Expand brackets with positive and negative factors
  • Handle a bare negative sign before brackets as multiplication by −1
  • Expand algebraic factors (e.g. 3x × term) using exponent addition
  • Expand multiple brackets and simplify by collecting like terms
  • Factorise expressions by extracting the highest common factor
  • Verify factorisations by re-expanding
Up Next → Lesson 13
Substitution

Replace variables with specific values and evaluate expressions — putting every skill from Stage I to work together.

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Lesson 11 - Collecting Like Terms
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Lesson 13 - Substitution