Stage 4 · Year 8 · MA4-ALG-C-01

Algebraic
Techniques

Algebra is one of mathematics' most powerful tools — a language for describing patterns, relationships and unknowns. In this unit you'll build your fluency in reading, writing and manipulating algebraic expressions.

7 Lessons
~4–5 Weeks
Mixed Ability
Unit Progress
0 of 7 complete

Quick Access

📋
Unit Pre-Quiz 7 questions — see what you already know
🌙
History Lens: al-Khwarizmi Meet the mathematician who invented algebra
✅
Unit Post-Quiz 7 questions — measure what you've learned

Lessons

1

Pronumerals & Algebraic Notation

What is a pronumeral? Reading and writing the conventions of algebra.

Ready
Notation Definitions
2

Substitution

Replacing pronumerals with values and evaluating expressions step by step.

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Substitution BIDMAS Connection Negatives
3

Generating Patterns

Using expressions as rules to produce number sequences; finding the rule from a table.

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Patterns Translation Finding Rules
4

Simplifying Expressions

Collecting like terms; commutative and associative properties in algebra.

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Like Terms Properties Collecting
5

Multiplying & Dividing Terms

Simplifying expressions with ×, ÷, algebraic fractions and mixed operations.

Ready
Multiplication Division Mixed Ops
6

Expanding Expressions

The distributive law and grouping symbols — multiplying through brackets.

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Distributive Law Brackets Expand & Simplify
7

Factorising Expressions

Finding HCF, factorising with numerical and algebraic factors — the reverse of expanding.

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HCF Factorising Final Lesson
1
Lesson 1 of 7

Pronumerals & Algebraic Notation

What is a pronumeral? Conventions for reading and writing algebraic expressions.

Define pronumeral & variable Define algebraic expression Read & use algebraic notation Explain conventions for ×, ÷, powers
📜
History Lens: Where does algebra come from?

Before we start — meet the 9th-century mathematician whose name gave us the word algebra.

→

🔎 You've already been doing algebra

Think back to the formulas you've used in mathematics and science. Every time you used a letter to stand for a measurement, you were using the language of algebra.

Here are some you've almost certainly seen before:

A = l × w   |   P = 2l + 2w   |   d = st
💬 Discuss with a partner

In the formula A = l × w, what does each letter represent? Can l be different values in different problems? What does that tell us about what letters can do in mathematics?

The letters in these formulas — A, l, w, d, s, t — are all standing in for numbers. They are placeholders. And that placeholder idea is the heart of algebra.

💡 A problem from ancient Baghdad

Around the year 820 CE, a mathematician named al-Khwarizmi described this kind of problem in his famous book:

"If from a number you subtract its third and its quarter, nine remains. What is the number?"

He had no letters or symbols — he described the unknown in words. Today we'd write it as: x − x/3 − x/4 = 9

💬 Think about it

Why might using a letter like x be more useful than describing the unknown in words? Can you think of a situation where having a symbol for "the unknown" makes a problem much easier to work with?

🎯
Learning Goal for this lesson By the end of this lesson you will be able to: define and use the terms pronumeral, variable and algebraic expression; and correctly read and write algebraic notation including conventions for multiplication, division and powers.

📖 Key Definitions

Pronumeral
A letter or symbol used to represent a numerical value. The word comes from pro (in place of) + numeral (number). A pronumeral is a placeholder for a number.
Variable
When a pronumeral can take more than one numerical value, it is called a variable. For example, in the formula A = l × w, the letters l and w are variables — they change depending on the rectangle.
🔑
All variables are pronumerals, but not all pronumerals are variables. When we write x = 5, the letter x is a pronumeral for a specific, fixed value. When we write y = 2x + 1 without fixing x, then x is a variable.
Algebraic Expression
An expression formed by combining numbers and algebraic symbols (pronumerals) using arithmetic operations (+, −, ×, ÷, powers). An algebraic expression does not have an equals sign — it is not an equation.
💡
Expression vs. Equation
3x + 2 is an expression — a combination of terms.
3x + 2 = 11 is an equation — it makes a statement that two things are equal.

✏️ Algebraic Notation Conventions

Algebra uses a set of agreed shorthand conventions to write expressions more concisely. These are not optional — they are the language rules of algebra that everyone agrees to follow.

Expression What it means Read as
ab a × b (multiplication sign omitted) "a times b" or "a b"
3x 3 × x (coefficient written first) "three x" or "three times x"
x/4 or x ÷ 4 x divided by 4 "x over four"
x² x × x (x multiplied by itself) "x squared"
x³ x × x × x "x cubed"
4(x + 2) 4 × (x + 2) "four times the quantity x plus two"
−x −1 × x (the negative of x) "negative x"
⚠️
Why no × sign? We drop the multiplication sign between a number and a letter (writing 3x not 3 × x) because the × symbol can be confused with the letter x. This convention also makes expressions much easier to read and write.

🔍 Worked Examples

Let's practise reading and writing algebraic expressions correctly.

Example 1 — Write as an algebraic expression
"Five more than three times a number n"
  • Identify the operations: three times n, then add five
  • Write "three times n" using convention: 3n
  • Add five: 3n + 5
  • Answer: 3n + 5
Example 2 — Describe in words
Describe the expression: 2x² − 7
  • 2x² means 2 times x squared
  • Then subtract 7
  • Answer: "Seven less than twice the square of x" (or "two times x squared, minus seven")
Example 3 — Identify the parts
In the expression 5a + 3b − 2, identify each term and explain the notation.
  • 5a means 5 × a (the coefficient of a is 5)
  • 3b means 3 × b (the coefficient of b is 3)
  • −2 is a constant term (no pronumeral)
  • The expression has 3 terms, 2 pronumerals (a and b) and uses addition and subtraction

🕵️ Error Detective

A student has made errors in the work below. Find the mistake and explain what the correct answer should be.

Error Detective #1 Notation Convention
Student's Working

Write "the product of 4 and x" as an algebraic expression.

Answer: 4 × x

Is this expression wrong? Why or why not?
✅ Explanation

The student's mathematical value is correct — 4 × x does equal the product of 4 and x. However, the notation is not in accepted algebraic form. In algebra, we omit the × sign between a coefficient and a pronumeral to avoid confusion with the letter x. The correct algebraic expression is 4x.

Error Detective #2 Reading a Power
Student's Working

Describe what x² means.

Answer: "x times 2" or "2x"

What is the error? What does x² actually mean?
✅ Explanation

This is a very common mistake. The small raised 2 in x² is a power (or index), not a multiplication. x² means x × x — "x squared" or "x multiplied by itself." It is completely different from 2x, which means 2 × x. For example, if x = 5: x² = 5 × 5 = 25, but 2x = 2 × 5 = 10.

Error Detective #3 Expression vs. Equation
Student's Working

Teacher asks: "Write an algebraic expression for double a number m, plus one."

Answer: 2m + 1 = 0

What has the student confused? What should the answer be?
✅ Explanation

The student has written an equation (with an = sign) when the question asked for an expression. An expression is just a combination of terms — it does not make a statement about equality. The correct answer is simply 2m + 1, with no equals sign.

🔗 Matching Activity

Match each algebraic expression (left) with its meaning in words (right). Click an expression, then click its matching description.

Algebraic Expression
5n + 3
n²
n/2
3(n + 4)
Meaning in Words
Half of n
Three more than five times n
Three times the sum of n and four
n multiplied by itself

⭐ Independent Practice

Choose your starting tier. Foundation builds the core idea; Core is the expected level; Extension pushes deeper.

🟢
Foundation — Focus on understanding what each piece of notation means. Use the notation table in the I Do tab if you need to.
1
Circle all of the algebraic expressions in this list (not equations):   3x + 1   y = 5   2a − 7   n² + n = 6   4p   m/3 + 2
2
Write what each expression means in plain words:
(a) 6k    (b) x + 9    (c) m²    (d) y/5
3
Write each of the following as an algebraic expression using correct notation:
(a) "A number n multiplied by 4"   (b) "A number p divided by 3"   (c) "A number q squared"
Hint: remember we write the coefficient before the pronumeral, and we don't need the × symbol
4
True or false? Explain your answer for each.
(a) x² and 2x always have the same value.
(b) 3ab means 3 × a × b.
(c) The expression 5n − 2 has two terms.
🔵
Core — Work through all questions. Show your reasoning, not just answers.
1
Explain in your own words the difference between a pronumeral and a variable. Give an example of each.
2
Write each description as an algebraic expression using correct notation:
(a) "Eight less than twice a number m"
(b) "The square of a number p, plus seven"
(c) "A number x divided by five, minus three"
(d) "The product of four, a and b"
3
Describe each expression in words:
(a) 3n² + 1    (b) 5(x − 2)    (c) 2a/3 + b
4
A student says: "In the expression 7y, the 7 and the y are both pronumerals." Is this correct? Explain why or why not, and give the correct terminology for each part.
5
The perimeter of a rectangle is given by P = 2l + 2w.
(a) Identify the pronumerals in this formula. Are they variables? Explain.
(b) Rewrite this formula using even more concise algebraic notation (hint: what can you do with the two separate 2s?)
🔴
Extension — These questions ask you to think more deeply about conventions, structure and mathematical reasoning.
1
Algebraic notation has evolved over centuries. Research (or reason about) why mathematicians eventually agreed to drop the × sign when multiplying a coefficient and a pronumeral. Why is 3x better notation than 3 × x in an algebraic context? What might go wrong if we kept the × sign?
2
Consider the two expressions 2 + 3x and 3x + 2.
(a) Are these the same expression? How can you be sure?
(b) What property of arithmetic justifies why the order of addition doesn't matter?
(c) Does the same apply to 2 × 3x versus 3x × 2?
This connects to what you'll learn about the commutative property in Lesson 4
3
How many different algebraic expressions can you write using only the pronumerals x and y and the numbers 2 and 3 (each used at most once)? List at least 8, and explain any patterns you notice about how they differ.
4
Al-Khwarizmi wrote his algebra problems entirely in words (rhetorical algebra) in 820 CE. It took another 700+ years before the symbolic notation we use today was standardised.
(a) Try writing the expression 3x² + 2x − 5 entirely in words, the way al-Khwarizmi might have.
(b) What are the advantages and disadvantages of symbolic notation versus verbal description?
(c) Can you think of a situation in modern life where important information is still written entirely in words rather than symbols, and consider what the tradeoffs are?
Baghdad · ~820 CE · House of Wisdom

History Lens: al-Khwarizmi

The mathematician whose work gave algebra its name — and changed how humanity thinks about unknowns.

Who was al-Khwarizmi?

Muhammad ibn Musa al-Khwarizmi was a scholar who worked in Baghdad around 820 CE, during the golden age of Islamic scholarship. He worked at the famous House of Wisdom (Bayt al-Hikma) — a great library and research centre where scholars from across the known world gathered to translate, study and advance human knowledge.

Al-Khwarizmi was a polymath: he made major contributions to mathematics, astronomy and geography. But his most lasting legacy is a single book that transformed mathematics forever.

The Book That Changed Mathematics

Around 830 CE, al-Khwarizmi wrote a book whose Arabic title was al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wal-muqābala — roughly translated as The Compendious Book on Calculation by Completion and Balancing.

The word al-jabr (الجبر) — meaning "completion" or "restoration" — gives us the English word algebra. Every time you write the word algebra, you are speaking a word from 9th-century Arabic.

The book systematically described methods for solving linear and quadratic equations — but entirely in words. Al-Khwarizmi had no symbols, no x or y, no equals sign. He described everything in complete sentences.

Al-Khwarizmi in Practice

Here is how al-Khwarizmi described what we would write as x² + 10x = 39:

"A square and ten roots equal thirty-nine dirhams. What is the square? You halve the number of roots, which gives five; multiply this by itself to get twenty-five; add this to thirty-nine to get sixty-four; take the square root, which is eight; subtract half the roots, five, to get three. This is the root of the square."

Read that carefully — he just solved a quadratic equation using only words and geometric reasoning, with no algebraic symbols at all. The entire procedure is the mathematics.

🤔 Think About It

Check his answer: if the root (solution) is x = 3, does x² + 10x = 39? What does this tell you about how powerful correct reasoning can be, even without modern notation?

Another Legacy: Algorithms

Al-Khwarizmi also wrote a book explaining the Hindu-Arabic numeral system (the digits 0–9 we use today) to the Arab world. When this book was later translated into Latin in medieval Europe, his name was Latinised as Algoritmi.

The systematic, step-by-step methods he described became associated with his name. By the 13th century, a "step-by-step procedure" had become known as an algorismus — from which we get the modern word algorithm.

When you run a Google search, stream a video, or ask a navigation app for directions, you are using algorithms. That word traces directly to a mathematician in 9th-century Baghdad.
🤔 Think About It

Two words used daily in modern mathematics and computing — algebra and algorithm — both trace back to one person. What does this tell you about how mathematical ideas spread across time and cultures?

From Words to Symbols: A 700-Year Journey

Al-Khwarizmi's algebra used no symbols — just carefully reasoned words. It took nearly 700 years for the symbolic notation we use today to develop:

  • ~1200s — Fibonacci introduces Hindu-Arabic numerals to Europe
  • ~1500s — Plus (+) and minus (−) signs appear in German texts
  • 1557 — Robert Recorde introduces the equals sign (=) in England
  • ~1600s — François Viète and René Descartes develop the use of letters for unknowns
  • ~1700s — Leibniz and Newton establish much of modern mathematical notation
🤔 Connecting to the Lesson

In Lesson 1, you are learning notation conventions that took centuries to develop. Why do you think the mathematical community bothered to develop and agree on these conventions? What would mathematics look like if every country or era used different symbols?

2
Lesson 2 of 7

Substitution

Replacing pronumerals with values and evaluating expressions correctly.

Use brackets when substituting Evaluate with positive integers Evaluate with negative integers Apply BIDMAS after substituting Substitute into formulas

🔌 Expressions as machines

In Lesson 1 you learned to read and write algebraic expressions. But an expression like 3x + 1 on its own doesn't have a value — x could be anything.

Think of an algebraic expression like a machine: you feed a number in, the machine applies the rule, and you get an output. Substitution is the act of feeding the number in.

Input
x = 4
Rule
3x + 1
Output
= 13
💬 Discuss

If the rule is 3x + 1, what output do you get when x = 0? When x = 10? When x = −2? What happens to the output as x gets larger?

📐 Why does this matter?

Substitution is how we use algebra in real life. Every formula you'll ever use — in science, finance, engineering, sport analytics, medicine — works by substituting real values and getting a result.

Real-world example — BMI formula
Body Mass Index uses the formula: BMI = w / h² where w = weight (kg), h = height (m). For a person weighing 72 kg and 1.80 m tall — what is their BMI?
  • Substitute: w = 72, h = 1.80
  • BMI = 72 / (1.80)²
  • BMI = 72 / 3.24
  • BMI ≈ 22.2
🔑
The key skill this lesson — Substitution is simple in concept, but easy to get wrong in execution. The most common errors come from forgetting notation conventions and rushing through order of operations. We'll look carefully at both.
🔗
Connecting to Lesson 1 — In Lesson 1 you learned that 3x means 3 × x, that x² means x × x, and that we always apply BIDMAS. Both of those facts are essential for evaluating expressions correctly. If you need a refresher on either, go back to Lesson 1's I Do tab.

📖 The Substitution Procedure

There is a reliable, three-step method for substituting that avoids most common errors:

  1. 1
    Rewrite the expression, replacing each pronumeral with its value inside brackets. The brackets are your protection against sign errors.
  2. 2
    Simplify the brackets — apply powers and evaluate anything inside grouping symbols first.
  3. 3
    Apply BIDMAS to evaluate the remaining expression fully.
⚠️
Always use brackets when substituting. Writing 3(−4) instead of 3−4 prevents a very common error. The bracket makes it clear you are multiplying, not subtracting.

🔍 Worked Examples — Positive Values

Example 1 — Single variable, multiple terms
Evaluate 4x − 3 when x = 5
  • Substitute (using brackets): 4(5) − 3
  • Multiply first (BIDMAS — M before S): 20 − 3
  • Subtract: = 17
Example 2 — Expression with a power
Evaluate x² + 3x when x = 4
  • Substitute: (4)² + 3(4)
  • Apply powers first: 16 + 3(4)
  • Multiply next: 16 + 12
  • Add: = 28
Example 3 — Two variables
Evaluate 2a + 5b − 1 when a = 3, b = 2
  • Substitute both variables: 2(3) + 5(2) − 1
  • Multiply: 6 + 10 − 1
  • Work left to right: = 15

🔍 Worked Examples — Negative Values

Negative substitutions are where most errors occur. The bracket rule is especially important here.

Example 4 — Negative value, coefficient
Evaluate 3x + 7 when x = −4
  • Substitute (brackets essential with negatives): 3(−4) + 7
  • Multiply: −12 + 7
  • Add: = −5
Example 5 — Negative value raised to a power
Evaluate x² when x = −3
  • Substitute: (−3)²
  • Apply power: (−3) × (−3)
  • Negative × negative = positive: = 9
Example 6 — Subtracting a negative
Evaluate 10 − 2x when x = −3
  • Substitute: 10 − 2(−3)
  • Multiply: 10 − (−6)
  • Subtracting a negative = adding: 10 + 6
  • = 16
💡
Negative squared vs negative of a square
(−3)² = 9 — the negative is inside the brackets, so we square −3
−3² = −9 — the negative is outside; only 3 is squared, then negated
The brackets make all the difference. Always substitute with brackets to be safe.

🔍 Worked Example — Substituting into a Formula

Example 7 — Area of a triangle
The area of a triangle is A = ½bh. Find A when b = 8 cm and h = 5 cm.
  • Substitute: A = ½ × (8) × (5)
  • Multiply: A = ½ × 40
  • = 20 cm²

🕵️ Error Detective

Each example below contains a substitution error. Identify the mistake and explain the correct working.

Error Detective #1 The invisible multiplication trap
Student's Working

Evaluate 3x + 2 when x = 4

3x + 2
= 34 + 2
= 36

Where is the error? What is the correct answer?
✅ Explanation

The student treated 3x as if the 3 and x were digits in a two-digit number, writing 34. But 3x means 3 × x, not "thirty-something." When x = 4: 3x = 3 × 4 = 12. Correct answer: 12 + 2 = 14. Using brackets helps: 3(4) + 2 = 12 + 2 = 14.

Error Detective #2 Negative value squared
Student's Working

Evaluate x² + 1 when x = −3

x² + 1
= −3² + 1
= −9 + 1
= −8

What went wrong with the power? What is the correct answer?
✅ Explanation

The student forgot to put brackets around the negative value before squaring. When we substitute x = −3, we must write (−3)² — meaning (−3) × (−3) = +9 (negative × negative = positive). Without brackets, −3² means −(3²) = −9, which is different. Correct: (−3)² + 1 = 9 + 1 = 10.

Error Detective #3 Missing brackets with a negative substitution
Student's Working

Evaluate 5 − x when x = −2

5 − x
= 5 − −2
= 5 − 2
= 3

Find the error in line 3. What is the correct answer?
✅ Explanation

In line 3, the student incorrectly simplified 5 − −2 as 5 − 2. But subtracting a negative is the same as adding: 5 − (−2) = 5 + 2 = 7. Using brackets when substituting makes this clearer: 5 − (−2). The two negatives together signal that the signs combine to give a positive.

Error Detective #4 BIDMAS breakdown after substitution
Student's Working

Evaluate 3x² − 4 when x = 2

3x² − 4
= 3(2)² − 4
= 6² − 4
= 36 − 4
= 32

Where did BIDMAS break down? What is the correct answer?
✅ Explanation

The student multiplied 3 × 2 before applying the power, getting 6². But in 3x², the power applies only to x, not to the 3. BIDMAS: powers before multiplication. So: 3(2)² = 3 × (2²) = 3 × 4 = 12. Correct answer: 12 − 4 = 8. If you wanted to square the whole of 3x, you'd write (3x)².

⚙️ Substitution Machine

Choose an expression, enter a value for x, and see each step worked out. Try different values — including negative ones.

x =

⭐ Independent Practice

Choose your tier. Remember to show full working — substitution with brackets, then BIDMAS step by step.

🟢
Foundation — Positive integers only. Focus on using brackets and showing every step. Use the I Do worked examples as a model.
1
Evaluate each expression when x = 3:
(a) 2x    (b) x + 7    (c) 5x − 4    (d) x²
Hint: write 2(3), then calculate
2
Evaluate each expression when n = 4:
(a) 3n + 1    (b) 10 − n    (c) n² + 2
3
The perimeter of a square is P = 4s where s is the side length. Find P when:
(a) s = 6 cm    (b) s = 10 cm    (c) s = 2.5 cm
4
Evaluate 2a + b when a = 5 and b = 3.
Substitute both variables, then use BIDMAS
🔵
Core — Work with positive and negative integers. Always show the substitution line (with brackets) and BIDMAS working.
1
Evaluate each expression when x = −2:
(a) 4x + 1    (b) x² − 3    (c) 3 − 5x    (d) 2x² + x
2
Evaluate 3a − 2b + 1 when:
(a) a = 4, b = 2    (b) a = −1, b = 3    (c) a = 0, b = −5
3
The temperature in Fahrenheit is given by F = 1.8C + 32 where C is the temperature in Celsius.
(a) Convert 25°C to Fahrenheit.
(b) Convert 0°C to Fahrenheit. Does the answer make sense?
(c) Convert −10°C to Fahrenheit.
4
For the expression x² − 4x + 3, evaluate when x = 1, x = 2, and x = 3. What do you notice about the result when x = 1 and x = 3?
5
Without calculating, predict whether 2x + 5 will give a positive or negative result when x = −4. Then check by substituting. Was your prediction correct?
🔴
Extension — Work with fractions, decimals, and reverse problems. Justify your reasoning.
1
Evaluate each expression when x = ½:
(a) 4x + 1    (b) x² + x    (c) 8x² − 2x + 3
2
Reverse substitution: Find the value of x in each case.
(a) 3x + 1 = 10 — what value of x gives this result?
(b) x² − 1 = 8 — find a positive value of x that works.
(c) 2x + 5 = −3 — what value of x gives this result?
You don't need formal equation-solving techniques yet — try values systematically
3
The expression n² − n + 41 was studied by the mathematician Euler. Evaluate it for n = 1, 2, 3, 4, 5 and 10. What do you notice about all the results? (This connects to prime numbers — look into it!)
4
Two students are given the expression −x².
Student A says: "When x = −3, the answer is 9."
Student B says: "When x = −3, the answer is −9."
Who is correct? Write a precise explanation of why, referring to the order of operations and bracket conventions. Then explain how writing (−x)² would change the answer.
3
Lesson 3 of 7

Generating Patterns

Using algebraic expressions as rules that produce sequences — and working backwards to find the rule.

Generate terms from an expression Translate descriptions ↔ expressions Find a rule from a table of values Use first differences to identify linear rules

🔺 A pattern problem

Look at this sequence of triangle arrangements made from matchsticks:

n = 1 3 sticks n = 2 5 sticks n = 3 7 sticks ? n = 4 ? sticks
💬 Notice and wonder

How many matchsticks does pattern 4 need? How many does pattern 10 need? How many does pattern 100 need? Could you answer the last question without drawing it — and how would you explain your method to someone else?

The sequence of matchstick counts is: 3, 5, 7, 9, … Each pattern uses 2 more sticks than the previous one, starting from 3. Algebra gives us a way to describe this with a single rule instead of counting step by step.

🔗 Connecting to Lesson 2

In Lesson 2, you substituted a single value into an expression. In this lesson, we substitute a sequence of values — n = 1, 2, 3, 4, 5 — to generate a pattern.

The letter n represents the position number (also called the term number). Each time we substitute a new position, we get the corresponding term value.

Position
n
1st
1
2nd
2
3rd
3
4th
4
🎯
Learning goal for this lesson — You will be able to: generate the terms of a number pattern from an algebraic rule; translate between verbal descriptions and algebraic expressions; and find the rule (expression) that describes a given table of values.

📖 Part 1 — Generating a sequence from a rule

If we're given the expression, we substitute n = 1, 2, 3, 4, 5 one at a time to build the sequence.

Example 1 — Linear rule
Generate the first 5 terms of the sequence with rule: 3n + 1
  • n = 1: 3(1) + 1 = 4
  • n = 2: 3(2) + 1 = 7
  • n = 3: 3(3) + 1 = 10
  • n = 4: 3(4) + 1 = 13
  • n = 5: 3(5) + 1 = 16
  • Sequence: 4, 7, 10, 13, 16, …   (increases by 3 each time)
Example 2 — Rule with a coefficient of 1 implied
Generate the first 5 terms of: n + 5
  • n = 1: 1 + 5 = 6
  • n = 2: 2 + 5 = 7
  • n = 3: 3 + 5 = 8
  • n = 4: 4 + 5 = 9
  • n = 5: 5 + 5 = 10
  • Sequence: 6, 7, 8, 9, 10, …   (increases by 1 each time)
🔑
Key observation: For a rule of the form an + b, the coefficient a tells you the common difference — how much the sequence increases each step. This is the pattern's "jump size."

📖 Part 2 — Translating between words and expressions

This is a two-way skill. You need to move fluently in both directions.

Description in wordsAlgebraic expression
Five more than the term numbern + 5
Double the term number, minus three2n − 3
The term number multiplied by itselfn²
Three less than four times the term number4n − 3
Half the term number, plus onen/2 + 1
⚠️
Watch the order! "Three less than n" means n − 3 (start with n, then subtract 3). "n less than three" means 3 − n (start with 3, then subtract n). These are different expressions with different values.

📖 Part 3 — Finding the rule from a table of values

When we're given a table, we work backwards to find the expression. The key tool is first differences — the gaps between consecutive terms.

Example 3 — Find the rule
n (position) 1 2 3 4 5
Term value 5 8 11 14 17
  • Find first differences: 8−5=3, 11−8=3, 14−11=3. The difference is constant at 3, so the rule is linear and has the form 3n + b
  • Substitute n = 1, term = 5:   3(1) + b = 5  →  b = 2
  • Check with n = 2: 3(2) + 2 = 8 ✓   Check n = 3: 3(3) + 2 = 11 ✓
  • Rule: 3n + 2
Example 4 — Decreasing sequence
n 1 2 3 4
Term 20 17 14 11
  • First differences: 17−20=−3, 14−17=−3. Constant difference of −3, so rule is −3n + b
  • Substitute n = 1, term = 20:   −3(1) + b = 20  →  b = 23
  • Check n = 2: −3(2) + 23 = 17 ✓
  • Rule: −3n + 23   (or equivalently: 23 − 3n)
💡
The method in summary:
1. Calculate first differences (consecutive terms subtracted) → this gives you the coefficient of n
2. Substitute one known pair (n, term) to find the constant b
3. Verify with a second pair

🕵️ Error Detective

Each example below contains a mistake in reasoning about patterns. Identify what went wrong.

Error Detective #1 Confusing position with value
Student's Working

Find the rule for: 4, 7, 10, 13, …

"The sequence goes up by 3 each time, so the rule is n + 3."

The student has correctly identified the common difference, but the rule is wrong. Why?
✅ Explanation

The student confused the common difference (3) with the coefficient of n. n + 3 would give 4, 5, 6, 7 — not 4, 7, 10, 13. The common difference of 3 tells us the coefficient is 3, so the rule starts as 3n + b. Substituting n = 1, term = 4: 3(1) + b = 4, so b = 1. The correct rule is 3n + 1. Check: 3(2)+1 = 7 ✓, 3(3)+1 = 10 ✓.

Error Detective #2 Word-to-expression translation error
Student's Working

Write an expression for: "Four less than triple the term number."

Answer: 4 − 3n

The student got the terms right but the order wrong. What is the correct expression?
✅ Explanation

"Four less than triple n" means: start with triple n (which is 3n), then subtract 4. So the expression is 3n − 4. The student reversed it, writing 4 − 3n, which means "triple n less than four" — a completely different value. Test with n = 2: correct answer is 3(2) − 4 = 2; student's answer gives 4 − 3(2) = −2. Order matters when translating "less than."

Error Detective #3 Off-by-one start
Student's Working

Generate the first 4 terms of 2n + 3.

n = 0: 2(0) + 3 = 3
n = 1: 2(1) + 3 = 5
n = 2: 2(2) + 3 = 7
n = 3: 2(3) + 3 = 9
Sequence: 3, 5, 7, 9

The sequence values are actually correct — so what is the conceptual error?
✅ Explanation

Unless stated otherwise, sequences typically start at n = 1 (the first term), not n = 0. The student has generated the right four values but labelled them as the 0th through 3rd terms. The first four terms starting at n = 1 are: 5, 7, 9, 11. Starting at n = 0 gives different position-value relationships and would change how we interpret any rule. Always check whether a question asks for the first term to be at n = 0 or n = 1.

Error Detective #4 Assuming every pattern is the same type
Student's Working

Find the rule for: 1, 4, 9, 16, 25, …

"First differences: 3, 5, 7, 9 — they aren't constant. The differences are going up by 2 each time, which means… the rule is probably 2n + something. Let me try 2n − 1: I get 1, 3, 5, 7, 9. That's not right either."

The student correctly noticed the differences aren't constant. What type of pattern is this, and what is the rule?
✅ Explanation

When first differences are not constant but second differences are constant (here the differences of differences are 2, 2, 2), the rule is quadratic — it involves n². The sequence 1, 4, 9, 16, 25 is the perfect squares: 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25. The rule is simply n². Not all patterns are linear — when first differences are non-constant, look for a quadratic (n²) pattern.

🔢 Pattern Generator

Enter an expression using n as the term number. The generator will substitute n = 1 through 6 and display the sequence with working.

Use: n, n^2 for n², * for multiply, +, -, /. Examples: 3*n+1, n^2-n, 2*n-5

Try:

⭐ Independent Practice

Choose your tier. Show your working for each step — substitutions, first differences, and rule verification.

🟢
Foundation — Focus on generating sequences from given rules and simple translations. Substitute n = 1, 2, 3, 4, 5 in each case.
1
Generate the first 5 terms of each sequence:
(a) n + 4    (b) 2n    (c) 5n − 2
Substitute n = 1, then 2, then 3, then 4, then 5 each time
2
Write an algebraic expression for each description:
(a) "Six more than the term number n"
(b) "Triple the term number"
(c) "The term number multiplied by four, minus one"
3
A sequence starts: 3, 6, 9, 12, 15, …
(a) What is the common difference?
(b) Describe the pattern in words.
(c) Explain why the rule is 3n.
Check: does 3(1) = 3? Does 3(2) = 6?
4
Use the rule 4n − 1 to find:
(a) The 6th term    (b) The 10th term    (c) The 20th term
You don't need to list all terms — just substitute the position number directly
🔵
Core — Generate sequences, find rules from tables, and translate both ways. Show the first differences working when finding rules.
1
Find the rule for each table of values. Show your first differences and verify your rule with two substitutions.

(a)

n1234
T7111519

(b)

n1234
T10864
2
The matchstick triangle pattern from the Ignition task follows a rule. The sequence is 3, 5, 7, 9, …
(a) Find the algebraic rule.
(b) How many matchsticks would you need for the 15th pattern?
(c) Which pattern number requires exactly 51 matchsticks?
For (c): set your expression equal to 51 and find n
3
For the rule 3n − 1:
(a) Generate the first 5 terms.
(b) Is the number 50 a term in this sequence? Explain how you know.
(c) Is the number 47 a term in this sequence? Find its position.
4
Two athletes are training. Athlete A runs 2n + 5 km in week n. Athlete B runs 3n + 1 km in week n.
(a) How far does each athlete run in week 4?
(b) In which week do they run the same distance?
(c) After week 1, which athlete is training harder each week? How can you tell from the expressions?
🔴
Extension — Investigate quadratic patterns, build proofs using algebra, and think about what expressions can and cannot represent.
1
The sequence 2, 5, 10, 17, 26, … has first differences 3, 5, 7, 9 — which are not constant. Calculate the second differences (differences of the differences). What do you notice?
(a) Explain why a constant second difference suggests the rule involves n².
(b) Find the rule. (Hint: try n² + 1 and check.)
The general form for a quadratic pattern is an² + bn + c
2
A square garden has side length n metres. It is surrounded by a border of 1-metre square tiles.
(a) Draw diagrams for n = 1, 2, 3 and count the border tiles.
(b) Find an algebraic rule for the number of border tiles in terms of n.
(c) Verify your rule for n = 4.
(d) Can you express your rule in more than one algebraic form? Show that the forms are equivalent.
Think about it as 4 sides of (n+2) tiles with overlapping corners removed
3
Prove algebraically (using your expression) that the sequence generated by 2n + 1 always produces odd numbers. Why does the structure of the expression guarantee this?
4
Two students disagree about a rule. The table shows:
n1234
T03815
Student A says the rule is n² − 1. Student B says the rule is n(n−1) + (n−1). Who is right? Are both right? Show using algebra whether the two expressions are equivalent.
4
Lesson 4 of 7

Simplifying Expressions

Collecting like terms and using the properties of arithmetic to simplify algebraic expressions.

Commutative property in algebra Associative property in algebra Identify like terms Add and subtract like terms Simplify multi-term expressions

🍎 You can only add like things

Think about this: if you have 3 apples and 2 apples, you have 5 apples. Simple. But what if you have 3 apples and 2 oranges?

🍎🍎🍎 + 🍎🍎
3 apples + 2 apples = 5 apples ✓
🍎🍎🍎 + 🍊🍊
3 apples + 2 oranges = ?
Can't combine — different types!

Algebra works the same way. You can only add or subtract terms that are the same type — that is, terms that have exactly the same pronumeral(s) raised to the same power(s). These are called like terms.

💬 Before we start — predict

Look at this expression:  3x + 2y + 5x − y + 4

Which terms do you think can be combined? How many separate groups can you see? Make a prediction before moving to the I Do tab.

🔁 Properties you already know — now in algebra

You know from arithmetic that the order of addition doesn't change the result, and that regrouping additions doesn't either. In Lesson 3's extension task you had a hint of this. Now we make it explicit.

Arithmetic example
3 + 5 = 5 + 3 = 8
(2 + 3) + 4 = 2 + (3 + 4) = 9
Does it still work?
3x + 5x = 5x + 3x?
(2x + 3x) + 4x = 2x + (3x + 4x)?
💬 Discuss

Try substituting x = 2 into both sides of each expression above. Do they give the same value? What does this tell you about whether these properties hold in algebra?

🎯
Learning goal — By the end of this lesson you will be able to: identify like terms in any algebraic expression; add and subtract like terms to simplify; and apply the commutative and associative properties to rearrange and regroup terms before collecting.

📖 The Properties of Arithmetic — Generalised

Commutative Property
The order of addition or multiplication does not affect the result.
In arithmetic:   3 + 5 = 5 + 3   and   3 × 5 = 5 × 3
In algebra:   a + b = b + a   and   ab = ba

This means we can reorder terms in an expression to group like terms together.
Associative Property
The way additions or multiplications are grouped does not affect the result.
In arithmetic:   (2 + 3) + 4 = 2 + (3 + 4)
In algebra:   (a + b) + c = a + (b + c)

This means we can regroup terms to make collecting easier.
⚠️
Subtraction and division are NOT commutative.
5 − 3 ≠ 3 − 5, and 8 ÷ 2 ≠ 2 ÷ 8.
In algebra: a − b ≠ b − a in general. Take care when reordering expressions that involve subtraction — the sign belongs to the term that follows it.

📖 Like Terms

Like Terms
Terms are like terms if they have exactly the same pronumeral(s) raised to exactly the same power(s). The coefficients may be different — only the algebraic part matters.
Like terms ✓Unlike terms ✗Reason
3x and 7x 3x and 7y Different pronumerals (x ≠ y)
5a² and −2a² 5a² and 5a Different powers (a² ≠ a)
4xy and −xy 4xy and 4x One term is missing the y factor
6 and −11 6 and 6x Constants can't combine with pronumeral terms
🔑
The key check: cover up the coefficients — if the remaining algebraic parts are identical, the terms are like terms. For example, 5x²y and −3x²y → covering coefficients gives x²y and x²y — identical, so they are like terms.

📖 Collecting Like Terms — Worked Examples

When we collect like terms, we add or subtract their coefficients and keep the algebraic part the same. Think of it like the fruit analogy: 3 apples + 5 apples = 8 apples.

Example 1 — One type of like term
Simplify: 5x + 3x
  • Both terms have the same algebraic part: x
  • Add coefficients: 5 + 3 = 8
  • = 8x
Example 2 — Subtracting like terms
Simplify: 9a − 4a
  • Same algebraic part: a
  • Subtract coefficients: 9 − 4 = 5
  • = 5a
Example 3 — Multiple groups of like terms
Simplify: 3x + 2y + 5x − y + 4
  • Use commutative property to group like terms: (3x + 5x) + (2y − y) + 4
  • Collect x terms: 3x + 5x = 8x
  • Collect y terms: 2y − y = 1y = y   (coefficient 1 is not written)
  • Constant: 4 (no like terms to collect)
  • = 8x + y + 4
Example 4 — Negative coefficients, including x²
Simplify: 4x² − 3x + x² + 7x − 2
  • Group: (4x² + x²) + (−3x + 7x) + (−2)
  • x² terms: 4x² + 1x² = 5x²   (remember: x² has coefficient 1)
  • x terms: −3x + 7x = 4x
  • Constant: −2
  • = 5x² + 4x − 2
💡
The invisible coefficient of 1 — x means 1x, x² means 1x², and −x means −1x. When collecting like terms, always include this hidden 1 in your count: 4x² + x² = 4x² + 1x² = 5x².
Example 5 — Sign belongs to the term after it
Simplify: 6m − 5n − 2m + 3n
  • The minus sign travels with the term it precedes: (6m − 2m) + (−5n + 3n)
  • m terms: 6m − 2m = 4m
  • n terms: −5n + 3n = −2n
  • = 4m − 2n

🕵️ Error Detective

Find and explain the mistake in each student's simplification.

Error Detective #1 Combining unlike terms
Student's Working

Simplify: 3x + 2y

= 5xy

What fundamental rule has the student broken? What is the correct simplified form?
✅ Explanation

3x and 2y are unlike terms — they have different pronumerals. They cannot be combined. The student incorrectly added the coefficients (3 + 2 = 5) and multiplied the pronumerals (x × y = xy). The expression 3x + 2y is already fully simplified — it cannot be written more concisely. Just as 3 apples + 2 oranges cannot be collapsed into 5 of anything, unlike terms cannot be merged.

Error Detective #2 Forgetting the invisible coefficient
Student's Working

Simplify: 5x + x

5x + x
= 5 + x     (taking x from both)
= 5 + x

The student "cancelled" the x. What did they forget? What is the correct answer?
✅ Explanation

The second term x has an invisible coefficient of 1 — it means 1x. You add the coefficients, not "cancel" the variables: 5x + 1x = (5 + 1)x = 6x. Think of it as 5 apples + 1 apple = 6 apples. You would never say "5 apples + 1 apple = 5 + apple."

Error Detective #3 Sign error when collecting
Student's Working

Simplify: 8x − 11x

= 3x

The student got a positive answer. Should it be positive? What is the correct answer?
✅ Explanation

The student subtracted 8 − 11 = −3 but then dropped the negative sign, writing 3x instead of −3x. When subtracting a larger coefficient from a smaller one, the result is negative: 8x − 11x = (8 − 11)x = −3x. A quick check by substitution: if x = 2, then 8(2) − 11(2) = 16 − 22 = −6. Does −3(2) = −6? Yes. Does 3(2) = 6? No.

Error Detective #4 Confusing powers with like terms
Student's Working

Simplify: 3x² + 2x

= 5x³

Two separate errors here — one with the type of terms, one with what happens to the index. Identify both.
✅ Explanation

Two errors: (1) 3x² and 2x are unlike terms — x² and x are different (just as apples and apple-pairs are different). They cannot be combined at all. (2) Even if they were like terms, collecting never adds the indices — you only add the coefficients and keep the algebraic part unchanged. Adding indices (2 + 1 = 3) is something that happens in multiplication, not addition. The expression 3x² + 2x is already fully simplified.

🧩 Like Terms Collector

Click on pairs of like terms in the expression below to highlight them in matching colours, then simplify. A new expression appears each round.

⭐ Independent Practice

Choose your tier. Show your grouping step before writing the simplified answer — don't skip straight to the result.

🟢
Foundation — Circle or underline like terms before collecting. Show the grouping step: (… + …) + (… + …).
1
Which of the following pairs are like terms? Write "like" or "unlike" for each.
(a) 4x and 9x    (b) 3a and 3b    (c) 5m² and 2m²    (d) 7 and 7x    (e) 6xy and −2xy
2
Simplify each expression:
(a) 4x + 3x    (b) 8a − 5a    (c) 6y + y    (d) 10p − 10p
Remember: y means 1y. What do you get when 10p − 10p = 0 × p?
3
Simplify by collecting like terms:
(a) 3x + 5 + 2x
(b) 4a + 3b + 2a
(c) 7m − 3m + 2n
Show your grouping: (3x + 2x) + 5 = ?
4
A rectangle has a length of 3x cm and a width of x cm. Write an expression for the perimeter and simplify it.
Perimeter = 2l + 2w. Substitute, then collect like terms.
🔵
Core — Show the grouping step and the collecting step as separate lines. Include sign checks.
1
Simplify fully:
(a) 5x + 3y − 2x + 4y
(b) 6a² + 3a − 2a² − 7a
(c) 4p + 3q − p − 5q + 2
(d) 8 − 3x + 4x − 1
2
Determine whether each pair of expressions is equivalent (simplifies to the same thing). Show full working.
(a) 3x + 4y − x and 2x + 4y
(b) 5a − 3b + 2a and 7a + 3b
(c) 4m − 2n + n − m and 3m − n
3
A student simplifies 7x − 3 − 4x + 8 and gets 11x + 5. Without simplifying yourself, explain how you could quickly check whether this answer is correct. Then check it.
4
Write an expression with at least 6 terms (using x, y and constants) that simplifies to 3x − 2y + 1. Then verify by collecting like terms.
5
A triangle has sides of length (3x + 1) cm, (x + 4) cm and (2x − 2) cm. Write and simplify an expression for the perimeter.
🔴
Extension — Justify using properties, prove using substitution, and reason about when expressions are or aren't equivalent.
1
Prove that a + b = b + a holds in algebra by choosing three different pairs of values for a and b and verifying. Then explain why this must always be true — not just for specific values, but for all possible values of a and b.
2
A student claims: "You can always simplify an expression by collecting like terms." Is this true? Give an example of an expression that cannot be simplified at all by collecting like terms, and explain why.
3
Consider the expression 3x + 5x².
(a) Explain why these are unlike terms.
(b) A student tries to prove they're unlike by substituting x = 1: "3(1) + 5(1²) = 8, and 8x² would give 8, so maybe they are like terms?" Identify the flaw in this reasoning.
(c) Find two values of x that give different results for 3x + 5x² compared to 8x², confirming they are not equivalent.
4
Subtraction is not commutative: a − b ≠ b − a in general.
(a) Show that 5x − 3x and 3x − 5x give different results.
(b) Under what specific condition would a − b = b − a be true?
(c) When we write 3x − 7y + 2x − y, we often reorder as (3x + 2x) + (−7y − y). Explain carefully why this reordering is valid — what property are we using, and what must we be careful about?
5
Lesson 5 of 7

Multiplying & Dividing Terms

Simplifying expressions involving multiplication, division, algebraic fractions, and mixed operations.

Multiply algebraic terms Divide algebraic terms Simplify algebraic fractions Apply BIDMAS to mixed operations

🔁 What you know — and what's different now

In Lesson 4 you added and subtracted like terms by changing the coefficients while keeping the algebraic part the same:

3x + 5x = 8x     (only the coefficients change)

Now we multiply and divide. The rule is fundamentally different: both the coefficients and the algebraic parts change.

💬 Before we start — predict

What do you think 3x × 2x equals? Write down a prediction. Is it 5x? 6x? 6x²? Something else? Keep your prediction in mind as you work through the I Do tab.

📦 An area model

Think of a rectangle with side lengths 3x and 2y. Its area is length × width:

Area = ?
3x
2y

Area = 3x × 2y. We multiply the coefficients together and the pronumerals together:

3 × 2 = 6    x × y = xy    →    3x × 2y = 6xy
💬 Extend the thinking

What if both sides were 3x? What would the area be then? Does your answer match your earlier prediction?

🎯
Learning goal — By the end of this lesson: multiply and divide algebraic terms by treating coefficients and variables separately; simplify algebraic fractions by cancelling common factors; and apply BIDMAS correctly when expressions involve mixed operations.

📖 Part 1 — Multiplying Algebraic Terms

When multiplying algebraic terms, multiply the coefficients together and the variable parts together — then write them combined.

🔑
The key principle: multiplication is commutative and associative, so we can rearrange freely. 3x × 4y = 3 × 4 × x × y = 12xy
Example 1 — Multiplying a term by a number
Simplify: 4 × 3x
  • Multiply coefficients: 4 × 3 = 12
  • Variable stays: x
  • = 12x
Example 2 — Multiplying two unlike terms
Simplify: 5a × 3b
  • Coefficients: 5 × 3 = 15
  • Variables: a × b = ab
  • = 15ab
Example 3 — Multiplying same-variable terms
Simplify: 3x × 4x
  • Coefficients: 3 × 4 = 12
  • Variables: x × x = x²   (x multiplied by itself)
  • = 12x²
Example 4 — Multiplying three terms
Simplify: 2x × 3y × 4
  • Coefficients: 2 × 3 × 4 = 24
  • Variables: x × y = xy
  • = 24xy
Example 5 — Multiplying with a negative
Simplify: −2a × 5b
  • Coefficients: −2 × 5 = −10
  • Variables: a × b = ab
  • = −10ab

📖 Part 2 — Dividing Algebraic Terms

Division works the same way in reverse: divide the coefficients separately from the variable parts.

Example 6 — Dividing a term by a number
Simplify: 12x ÷ 4
  • Divide the coefficient: 12 ÷ 4 = 3
  • Variable stays: x
  • = 3x
Example 7 — Dividing terms sharing a variable
Simplify: 15xy ÷ 3y
  • Write as a fraction: 15xy / 3y
  • Cancel common factor of 3: (15÷3) = 5
  • Cancel common factor of y: y/y = 1
  • = 5x
Example 8 — Same variable divided
Simplify: 10x² ÷ 2x
  • Write as fraction: 10x² / 2x
  • Coefficients: 10 ÷ 2 = 5
  • Variables: x² ÷ x = x   (x × x divided by x leaves one x)
  • = 5x

📖 Part 3 — Simple Algebraic Fractions

An algebraic fraction has a pronumeral in the numerator or denominator. To simplify, cancel any common factors between numerator and denominator.

Example 9 — Fraction with a single term
Simplify: 8x / 4
  • Divide coefficient by 4: 8 ÷ 4 = 2
  • = 2x
Example 10 — Fraction with a binomial numerator
Simplify: (6x + 9) / 3
  • Every term in the numerator is divided by 3
  • 6x ÷ 3 = 2x
  • 9 ÷ 3 = 3
  • = 2x + 3
⚠️
Divide every term, not just one. When a fraction has a sum in the numerator, the denominator divides all terms:   (6x + 9) / 3 = 2x + 3.   A common error is to write (6x + 9) / 3 = 6x + 3, dividing only the 9. Think of it as: the fraction bar groups the numerator the same way brackets do.

📖 Part 4 — Mixed Operations: BIDMAS Still Applies

When an expression contains both multiplication/division and addition/subtraction, BIDMAS determines the order. Multiply and divide before you add or subtract.

Example 11 — Multiply before adding
Simplify: 3x + 2 × 4x
  • BIDMAS: multiplication first → 2 × 4x = 8x
  • Now add like terms: 3x + 8x
  • = 11x
Example 12 — Full simplification with mixed operations
Simplify: 5x − 3x × 2 + 4x
  • Multiplication first: 3x × 2 = 6x
  • Rewrite: 5x − 6x + 4x
  • Collect like terms left to right: (5 − 6 + 4)x
  • = 3x
Example 13 — Multiple multiplications
Simplify: 2a × 3b + 4a × b
  • Multiply first term pair: 2a × 3b = 6ab
  • Multiply second term pair: 4a × b = 4ab
  • Now collect like terms: 6ab + 4ab = 10ab

🕵️ Error Detective

Find and explain the mistake in each student's working.

Error Detective #1 Adding instead of multiplying variables
Student's Working

Simplify: 3x × 2x

Coefficients: 3 × 2 = 6
Variables: x + x = 2x
= 6 × 2x = 12x

The student correctly multiplied the coefficients but mishandled the variables. What is the correct answer?
✅ Explanation

When multiplying variables, we multiply them — we don't add them. x × x = x² (x multiplied by itself), not x + x = 2x. The correct working: 3x × 2x = (3 × 2) × (x × x) = 6 × x² = 6x². A quick check: if x = 2, then 3(2) × 2(2) = 6 × 4 = 24. Does 6x² = 6(4) = 24? ✓ Does 12x = 12(2) = 24? In this case yes, but try x = 3: 3(3)×2(3) = 9×6 = 54, 6(9) = 54 ✓, 12(3) = 36 ✗.

Error Detective #2 Dividing coefficient but keeping the wrong variable
Student's Working

Simplify: 10x² ÷ 5x

10 ÷ 5 = 2, so the answer is 2x²

The student divided the coefficients correctly. What did they do wrong with the variable part?
✅ Explanation

The student divided the coefficients (10 ÷ 5 = 2) correctly but forgot to divide the variable part. We must also divide x² by x: x² ÷ x = x (since x² = x × x, dividing by x leaves one x). The full working: 10x² ÷ 5x = (10 ÷ 5) × (x² ÷ x) = 2 × x = 2x. Check with x = 3: 10(9) ÷ 5(3) = 90 ÷ 15 = 6. 2x = 6 ✓. 2x² = 18 ✗.

Error Detective #3 Dividing only one term in the numerator
Student's Working

Simplify: (10x + 15) / 5

Only the 15 is divisible by 5:
= 10x + 3

What rule did the student forget? What is the correct simplified form?
✅ Explanation

When dividing a sum by a number, every term in the numerator must be divided. The student divided 15 by 5 but left 10x unchanged. Correct working: (10x + 15) / 5 = 10x/5 + 15/5 = 2x + 3. Think of the fraction bar as brackets: (10x + 15) ÷ 5, meaning the 5 applies to the whole group, not just the last term.

Error Detective #4 BIDMAS ignored in mixed operations
Student's Working

Simplify: 6x + 3x × 2

= 6x + 3x × 2
= 9x × 2     (added 6x + 3x first)
= 18x

What order-of-operations rule was broken? What is the correct answer?
✅ Explanation

BIDMAS requires multiplication before addition. The student added 6x + 3x = 9x first, then multiplied by 2. The correct order: multiply 3x × 2 = 6x first, then add: 6x + 6x = 12x. Check with x = 1: 6(1) + 3(1) × 2 = 6 + 6 = 12. 12x = 12 ✓. 18x = 18 ✗.

⚙️ Step-by-Step Simplifier

Work through a mixed-operations expression one step at a time. Click Next Step to reveal each stage of the BIDMAS process.

Choose:

⭐ Independent Practice

Choose your tier. Always show the multiplication/division step first when using mixed operations — don't try to do it all at once.

🟢
Foundation — Focus on one operation at a time. For each multiplication: (coefficient × coefficient)(variable × variable). For each division: divide coefficients then divide variables.
1
Simplify each multiplication:
(a) 3 × 5x    (b) 4a × 2    (c) 2x × 3y    (d) 5p × 4p
2
Simplify each division:
(a) 8x ÷ 2    (b) 15a ÷ 5    (c) 12xy ÷ 4x    (d) 6m² ÷ 2m
3
Simplify each algebraic fraction:
(a) 10x / 5    (b) 9ab / 3    (c) (4x + 8) / 2
For (c): divide each term in the numerator by 2
4
A rectangle has sides 5x cm and 3 cm. Write and simplify an expression for its area.
🔵
Core — Show each step. For mixed operations, label which operation you are doing first.
1
Simplify fully. Show all working.
(a) 4x + 3 × 2x
(b) 6ab − 2a × b
(c) 3m × 2m + 4m²
(d) 10y ÷ 2 + 3y × 4
2
Simplify each algebraic fraction:
(a) (12x + 8) / 4
(b) (15a − 10) / 5
(c) 6x²y / 3x
3
Write expressions for each area and simplify:
(a) A square with side length 3x
(b) A rectangle with sides 2a and 5b
(c) A rectangle with sides 4x and 2x
4
A student says 3x + 4x × 2 and (3x + 4x) × 2 are the same thing. Are they? Calculate both and explain why the brackets in the second expression change the answer.
5
Simplify:   2a × 3b + 4a × b − a × 2b
Multiply all three products first, then collect the like terms
🔴
Extension — Multi-step problems, algebraic fractions with variables in the denominator, and proofs using substitution.
1
Simplify fully:
(a) 3x × 2x + 4x × x − x²
(b) (6a² + 9a) / 3a
(c) 2xy × 3x − 4x²y + xy × x
For (b): divide every term in the numerator by 3a
2
A rectangular pool has length 3x m and width 2x m. A tiled border of width 1 m surrounds it.
(a) Write an expression for the total area (pool + border).
(b) Subtract the pool area to find the area of the border alone.
(c) Simplify your answer and verify it with x = 5.
3
Show algebraically that a × b = b × a for all values by using specific numbers, then explain why this must be true using the commutative property of multiplication. Does this mean a ÷ b = b ÷ a? Investigate with examples.
4
Consider the expression n × (n + 1).
(a) Expand by multiplying out (this previews Lesson 6 — try to reason through it).
(b) Generate the first 6 values of this expression (n = 1, 2, 3, 4, 5, 6). What kind of numbers are these?
(c) Use algebra to explain why n(n+1) always produces an even number for any positive integer n.
6
Lesson 6 of 7

Expanding Expressions

The distributive law — multiplying through brackets to remove grouping symbols.

Explain the role of grouping symbols Apply the distributive law Expand with negative terms Expand with a variable factor Expand and simplify

🧮 You already know this law — from arithmetic

Mental multiplication often uses a strategy you may not have named. How would you calculate 6 × 23 mentally?

Two ways to calculate 6 × 23:

Standard
6 × 23 = 138
Split & distribute
6 × (20 + 3)
= 6×20 + 6×3
= 120 + 18 = 138

The second approach is the distributive law — multiplying the outside number across every term inside the brackets. In algebra, we use exactly the same idea:

6(20 + 3) = 6×20 + 6×3   ↔   a(b + c) = ab + ac
💬 Discuss

Use the "split and distribute" method to mentally calculate:   7 × 48,   5 × 97,   4 × 103. What split made each one easiest? How does this connect to what you're about to do with algebra?

📐 An area interpretation

The distributive law has a beautiful geometric meaning. Consider a rectangle with width a and total length (b + c):

ab ac a b c b + c = a(b + c) = ab + ac

Both rectangles together have the same area as the single large rectangle. This is why the law always works — it's a geometric truth.

🎯
Learning goal — By the end of this lesson: explain what grouping symbols mean in an algebraic expression; apply the distributive law to expand expressions including those with negative and variable factors; and expand then simplify by collecting like terms.

📖 Part 1 — Grouping Symbols

Grouping Symbols
Brackets (parentheses) in an algebraic expression indicate that the terms inside are treated as a single quantity. When a factor appears immediately before a bracket — with no operation sign — it means multiply.

3(x + 2) means 3 × (x + 2) — multiply the entire bracket by 3.
2x(x + 4) means 2x × (x + 4) — multiply the entire bracket by 2x.
🔑
An expression with brackets and one without brackets can look different but be equivalent — they give the same value for any substitution. Expanding removes the brackets to reveal the equivalent form:   3(x + 2) = 3x + 6. Both give the same result for any value of x.

📖 Part 2 — The Distributive Law

Distributive Law
a(b + c) = ab + ac

The factor outside the bracket is distributed (multiplied) across every term inside. This works for any number of terms inside the bracket and for subtraction:   a(b − c) = ab − ac
Example 1 — Positive integer factor
Expand: 4(x + 3)
  • Multiply 4 by the first term: 4 × x = 4x
  • Multiply 4 by the second term: 4 × 3 = 12
  • = 4x + 12
Example 2 — Subtraction inside the bracket
Expand: 5(2x − 3)
  • Multiply 5 by 2x: 5 × 2x = 10x
  • Multiply 5 by −3: 5 × (−3) = −15
  • = 10x − 15
Example 3 — Three terms inside the bracket
Expand: 3(2x + y − 4)
  • Multiply 3 by every term: 3×2x, 3×y, 3×(−4)
  • = 6x + 3y − 12
  • = 6x + 3y − 12

📖 Part 3 — Negative Factor Outside the Bracket

When the factor outside is negative, every sign inside flips. This is the most error-prone case.

Example 4 — Negative integer factor
Expand: −2(x + 5)
  • Multiply −2 by x: (−2) × x = −2x
  • Multiply −2 by 5: (−2) × 5 = −10
  • = −2x − 10
Example 5 — Negative factor, subtraction inside
Expand: −3(2x − 4)
  • Multiply −3 by 2x: (−3) × 2x = −6x
  • Multiply −3 by −4: (−3) × (−4) = +12   (negative × negative = positive)
  • = −6x + 12
⚠️
The sign rule for a negative factor:
negative × positive → negative     e.g. −3(x) = −3x
negative × negative → positive   e.g. −3(−4) = +12
A negative outside the bracket flips the sign of every term inside.

📖 Part 4 — Variable Factor Outside the Bracket

Example 6 — Pronumeral as the factor
Expand: x(x + 3)
  • Multiply x by x: x × x = x²
  • Multiply x by 3: x × 3 = 3x
  • = x² + 3x
Example 7 — Term with coefficient as the factor
Expand: 2x(3x − 5)
  • Multiply 2x by 3x: 2x × 3x = 6x²
  • Multiply 2x by −5: 2x × (−5) = −10x
  • = 6x² − 10x

📖 Part 5 — Expand and Simplify

When an expression contains two or more brackets, expand each one then collect like terms.

Example 8 — Two brackets added
Expand and simplify: 3(x + 2) + 2(x − 1)
  • Expand the first bracket: 3(x + 2) = 3x + 6
  • Expand the second bracket: 2(x − 1) = 2x − 2
  • Write together: 3x + 6 + 2x − 2
  • Collect like terms: (3x + 2x) + (6 − 2)
  • = 5x + 4
Example 9 — Brackets subtracted (critical case)
Expand and simplify: 4(x + 3) − 2(x + 5)
  • Expand first bracket: 4(x + 3) = 4x + 12
  • Expand second bracket — the minus sign distributes: −2(x + 5) = −2x − 10
  • Write together: 4x + 12 − 2x − 10
  • Collect like terms: (4x − 2x) + (12 − 10)
  • = 2x + 2
💡
The subtracted bracket trap: In 4(x + 3) − 2(x + 5), the minus sign in front of 2 is part of the factor — it's −2(x + 5), which gives −2x − 10. Never write just −(2x + 10) or accidentally give the second bracket a + result.

🕵️ Error Detective

Each example below contains an expansion error. Identify it and give the correct working.

Error Detective #1 Only distributing to the first term
Student's Working

Expand: 3(x + 4)

= 3x + 4

The student multiplied 3 by x but not by 4. Why must the factor multiply every term? What is the correct answer?
✅ Explanation

The bracket groups x + 4 as a single quantity. Multiplying by 3 means multiplying the entire group — every term inside. The distributive law requires: 3 × x AND 3 × 4. Correct expansion: 3x + 12. Check with x = 2: 3(2 + 4) = 3(6) = 18. Does 3x + 12 give 18 when x = 2? 6 + 12 = 18 ✓. Does 3x + 4? 6 + 4 = 10 ✗.

Error Detective #2 Sign error with negative factor
Student's Working

Expand: −4(x − 3)

−4 × x = −4x
−4 × −3 = −12
= −4x − 12

The student got the first term right. What went wrong with the second?
✅ Explanation

The student forgot the rule: negative × negative = positive. (−4) × (−3) = +12, not −12. The correct expansion is −4x + 12. A useful check: if x = 5, then −4(5 − 3) = −4(2) = −8. Does −4x + 12 = −20 + 12 = −8? ✓. Does −4x − 12 = −20 − 12 = −32? ✗. When a negative factor meets a negative term inside the bracket, the result is always positive.

Error Detective #3 Forgetting to expand before collecting
Student's Working

Expand and simplify: 2(x + 3) + 4x

= 2(5x + 3)   (added 4x inside the bracket)
= 10x + 6

The student moved 4x inside the bracket before expanding. Why is this wrong? What is the correct answer?
✅ Explanation

The 4x sits outside the bracket — it cannot be moved inside and added to x without changing the expression's value. The bracket must be expanded first: 2(x + 3) = 2x + 6. Then combine with the remaining terms: 2x + 6 + 4x = 6x + 6. Check with x = 1: 2(1 + 3) + 4(1) = 8 + 4 = 12. Does 6x + 6 = 12? ✓. Does 10x + 6 = 16? ✗.

Error Detective #4 Losing the negative when subtracting a bracket
Student's Working

Expand and simplify: 5(x + 2) − 3(x − 1)

= 5x + 10 − 3x − 3
= 2x + 7

The first bracket is correct. Find the error in expanding the second bracket.
✅ Explanation

When expanding −3(x − 1), both terms must be multiplied by −3: (−3)(x) = −3x and (−3)(−1) = +3, not −3. The student wrote −3 instead of +3 for the second term. Correct expansion: 5x + 10 − 3x + 3. Collecting: (5x − 3x) + (10 + 3) = 2x + 13. Check with x = 0: 5(2) − 3(−1) = 10 + 3 = 13. 2(0) + 13 = 13 ✓. 2(0) + 7 = 7 ✗.

📐 Area Model Expander

Choose an expression to expand. The area model shows how the factor outside distributes across each term inside. Watch each rectangle light up as the multiplication happens.

Choose:

⭐ Independent Practice

Choose your tier. Always show the distribution step — write out each multiplication before simplifying. This makes errors visible and fixable.

🟢
Foundation — Positive integer factors only. Write each multiplication step: factor × first term, then factor × second term. Verify one answer by substituting x = 2.
1
Expand each expression:
(a) 2(x + 5)    (b) 3(x − 4)    (c) 5(2x + 1)    (d) 4(3x − 2)
2
Expand and verify one answer by substituting x = 3:
(a) 6(x + 2)    (b) 4(2x − 3)
3
Expand and then simplify by collecting like terms:
(a) 2(x + 3) + 4x
(b) 3(x + 1) + 2(x + 4)
Expand both brackets first, then collect. Don't try to do it all in one step.
4
A rectangular garden has width 4 m and length (x + 3) m. Write an expanded expression for its area.
🔵
Core — Include negative factors and variable factors. Always expand before collecting. Show each bracket's expansion on its own line.
1
Expand each expression:
(a) −3(x + 2)    (b) −5(2x − 4)    (c) x(x + 7)    (d) 3x(2x − 5)
2
Expand and simplify fully:
(a) 3(x + 4) + 2(x − 1)
(b) 5(2x − 3) − 4(x − 2)
(c) 2(x² + 3x) − x(x − 1)
3
A rectangle has width 3x and length (2x + 5).
(a) Write an expression for its area and expand it.
(b) If x = 4, calculate the area using your expanded expression. Then verify by substituting x = 4 into the original factored form.
4
Show that 2(3x + 1) + 4(x − 2) and 10x − 6 are equivalent by expanding and simplifying the left side.
5
A student writes: "−(x + 3) = −x + 3 because you just put a negative in front." Is this correct? Expand −(x + 3) properly and explain the error in the student's reasoning.
🔴
Extension — Prove equivalences, work backwards from expanded form, and connect expansion to factorisation — a preview of Lesson 7.
1
Expand and simplify:
(a) x(x + 1) + x(x − 1)
(b) 3x(x + 2) − 2x(x + 3)
(c) a(a + b) + b(a + b)
For (c): expand both brackets, collect, then look carefully at what you can factor out of the result — this is the link to Lesson 7
2
Reverse engineering: An expression has been expanded to give 6x + 10. Write three different factored expressions that could have produced this result. What determines which factored form is considered the "fully factored" form?
3
Use the distributive law to prove that the sum of any two consecutive even numbers is always divisible by 4.
Let the smaller even number be 2n. What is the next consecutive even number? Add them and expand.
4
Consider: (x + 2)(x + 3). This involves two brackets multiplied together — which goes beyond our single-bracket work. Using the distributive law twice (treat (x + 2) as the factor and distribute it across x and 3), try to expand this product. What do you notice about the result?
Think of it as (x + 2) × x + (x + 2) × 3, then expand each
7
Lesson 7 of 7 — Final Lesson

Factorising Expressions

The reverse of expanding — finding the highest common factor and writing expressions in bracket form.

Find factors of algebraic terms Find the HCF of two or more terms Factorise with a numerical HCF Factorise with an algebraic HCF Verify by expansion

↩️ The reverse journey

In Lesson 6 you expanded expressions — multiplying a factor through a bracket to remove it. Factorising is exactly the reverse process: you start with the expanded form and find the bracket form.

Factorised form
3(x + 2)
Expand →
⇌
← Factorise
Expanded form
3x + 6

Both forms are equivalent — they give the same value for any substitution. Factorised form is often more useful: it reveals the structure of the expression, shows what divides into it evenly, and is essential for solving equations later in Stage 4 and Stage 5.

💬 Think first

Look at 12x + 8. What whole number divides into both 12 and 8? How would you write that as a factor in front of a bracket? What would be left inside? Make a prediction before moving to the I Do tab.

🔗 Connecting to what you already know

Factorising numbers is something you've done before. The factors of 12 are: 1, 2, 3, 4, 6, 12. Finding the highest common factor (HCF) of two numbers means finding the largest factor they share.

Numeric HCF (familiar)
HCF(12, 8) = 4
12 + 8 = 4(3 + 2) = 4(5) = 20
Algebraic HCF (new)
HCF(12x, 8) = 4
12x + 8 = 4(3x + 2)
🎯
Learning goal — By the end of this lesson: list factors of algebraic terms; find the HCF of two or more algebraic terms (numerical and algebraic parts); fully factorise expressions; and verify any factorisation by expanding back.

📖 Part 1 — Factors of an Algebraic Term

Just as a number has factors (divisors that leave no remainder), an algebraic term has factors too — all the terms that divide into it evenly.

Example 1 — Factors of a term
List all the factors of 12x²
  • Numerical factors of 12: 1, 2, 3, 4, 6, 12
  • Variable factors of x²: 1, x, x²
  • Combine each numerical factor with each variable factor:
  • Factors: 1, 2, 3, 4, 6, 12, x, 2x, 3x, 4x, 6x, 12x, x², 2x², 3x², 4x², 6x², 12x²
🔑
The factors of an algebraic term are all possible combinations of the factors of its coefficient with the factors of its variable part (including 1 and the term itself).

📖 Part 2 — Finding the HCF of Algebraic Terms

The HCF of two algebraic terms is found by taking the HCF of the coefficients and the HCF of the variable parts separately, then combining them.

Example 2 — HCF of two terms (numerical only)
Find the HCF of 6x and 10
  • HCF of coefficients: HCF(6, 10) = 2
  • Variable part: 10 has no variable, so no shared variable factor
  • HCF = 2
Example 3 — HCF with shared variable
Find the HCF of 6x² and 4x
  • HCF of coefficients: HCF(6, 4) = 2
  • HCF of variable parts: HCF(x², x) = x   (the lower power)
  • HCF = 2x
Example 4 — HCF of three terms
Find the HCF of 8x², 12x and 4
  • HCF of coefficients: HCF(8, 12, 4) = 4
  • Variable part: the last term (4) has no variable, so no shared variable
  • HCF = 4
💡
Variable HCF rule: For the variable part, take the lowest power that appears across all terms. HCF(x³, x², x) = x¹ = x. If any term has no variable at all, the variable part of the HCF is 1 (no variable).

📖 Part 3 — Factorising Using the HCF

To factorise, write the HCF outside a bracket and divide each original term by the HCF to get the terms inside the bracket.

Example 5 — Numerical HCF
Factorise: 6x + 10
  • HCF(6, 10) = 2   (no shared variable)
  • Write 2 outside: 2( )
  • Divide each term by 2: 6x ÷ 2 = 3x,   10 ÷ 2 = 5
  • = 2(3x + 5)
  • Verify: 2(3x + 5) = 6x + 10 ✓
Example 6 — Algebraic HCF (variable factor)
Factorise: x² + 5x
  • HCF(x², x) = x   (coefficients are both 1, lowest variable power is x)
  • Write x outside: x( )
  • Divide each term: x² ÷ x = x,   5x ÷ x = 5
  • = x(x + 5)
  • Verify: x(x + 5) = x² + 5x ✓
Example 7 — Combined numerical and algebraic HCF
Factorise: 6x² + 4x
  • HCF of coefficients: HCF(6, 4) = 2
  • HCF of variables: HCF(x², x) = x
  • HCF = 2x
  • Divide each term: 6x² ÷ 2x = 3x,   4x ÷ 2x = 2
  • = 2x(3x + 2)
  • Verify: 2x(3x + 2) = 6x² + 4x ✓
Example 8 — Three-term expression
Factorise: 6x² + 9x − 3
  • HCF of coefficients: HCF(6, 9, 3) = 3
  • Variable: the last term (−3) has no variable → HCF of variable parts = 1
  • HCF = 3
  • Divide each term: 6x² ÷ 3 = 2x²,   9x ÷ 3 = 3x,   −3 ÷ 3 = −1
  • = 3(2x² + 3x − 1)
  • Verify: 3(2x² + 3x − 1) = 6x² + 9x − 3 ✓
⚠️
Always find the HIGHEST common factor. Factorising 6x + 10 as 2(3x + 5) is correct, but only if 2 is truly the HCF. If you factor out something smaller — e.g. writing 12x + 8 = 2(6x + 4) instead of 4(3x + 2) — you have partially factorised but not fully. The expression inside the bracket should have no further common factors.

📖 Part 4 — Always Verify by Expansion

The single most reliable check for factorising is to expand your answer. If you get back the original expression, you are correct.

Example 9 — Full verify cycle
Factorise 10x² − 15x, then verify.
  • HCF(10, 15) = 5,   HCF(x², x) = x → HCF = 5x
  • 10x² ÷ 5x = 2x,   −15x ÷ 5x = −3
  • Factorised: 5x(2x − 3)
  • Verify by expanding: 5x(2x − 3) = 5x × 2x + 5x × (−3) = 10x² − 15x ✓

🕵️ Error Detective

Find and explain the error in each student's factorisation. Check by expanding.

Error Detective #1 Not finding the highest common factor
Student's Working

Factorise: 12x + 8

HCF = 2
= 2(6x + 4)

The student has factorised, but not fully. What is the actual HCF, and what is the complete factorisation?
✅ Explanation

The student factored out 2, but HCF(12, 8) = 4, not 2. Check: 2(6x + 4) expands back correctly, but the expression inside the bracket (6x + 4) still has a common factor of 2 — meaning the factorisation is incomplete. A fully factorised expression has no remaining common factor inside the bracket. Correct: 4(3x + 2). Verify: 4(3x + 2) = 12x + 8 ✓. Always check: can the terms inside the bracket be factorised further?

Error Detective #2 Missing the algebraic part of the HCF
Student's Working

Factorise: 6x² + 4x

HCF of 6 and 4 is 2
= 2(3x² + 2x)

The numerical HCF is correct. What did the student overlook? What is the fully factorised form?
✅ Explanation

Both terms also share a variable factor: x² and x both contain x. The HCF is not just 2 but 2x. Check: 2(3x² + 2x) expands correctly, but inside the bracket, 3x² and 2x still share a factor of x — so it's only partially factorised. Fully factorised: 2x(3x + 2). Verify: 2x(3x + 2) = 6x² + 4x ✓. The rule: always check both the coefficient HCF and the variable HCF.

Error Detective #3 Dividing only the coefficient, not the full term
Student's Working

Factorise: 6x + 9

HCF = 3
6 ÷ 3 = 2x,   9 ÷ 3 = 3
= 3(2x + 3)

The answer here actually happens to be correct — but look at the working for the first term. Is "6 ÷ 3 = 2x" the right way to show this step? What should it say?
✅ Explanation

The answer 3(2x + 3) is correct, but the working is misleading. The student wrote "6 ÷ 3 = 2x" — dividing just the coefficient and somehow producing the full term. The correct working should be: 6x ÷ 3 = 2x (dividing the whole term 6x by 3). In this case the variable x wasn't part of the HCF so it carries through untouched — but the working should reflect dividing the full term: 6x ÷ 3 = 2x. This matters more when the HCF includes a variable, e.g. 6x² ÷ 2x = 3x (not 3x²).

Error Detective #4 Sign error inside the bracket
Student's Working

Factorise: 10x² − 15x

HCF = 5x
10x² ÷ 5x = 2x,   −15x ÷ 5x = 3
= 5x(2x + 3)

Expand the student's answer to reveal the sign error. What is the correct factorised form?
✅ Explanation

Expanding the student's answer: 5x(2x + 3) = 10x² + 15x — but the original was 10x² − 15x. The sign was lost: (−15x) ÷ 5x = −3, not +3. Negative divided by positive = negative. Correct factorisation: 5x(2x − 3). Verify: 5x(2x − 3) = 10x² − 15x ✓. This is exactly why expanding to verify is essential — a sign error in the bracket is invisible until you check.

🧩 HCF Finder

Find the HCF of the two terms shown, then select the fully factorised expression. A new pair of terms appears each round.

Factorise:
Choose the correct factorised form:

⭐ Independent Practice

Choose your tier. Always verify each factorisation by expanding — write the check as part of your working.

🟢
Foundation — Numerical HCF only. After each factorisation, expand your answer to check. Show: HCF → bracket → verify.
1
List all factors of each term:
(a) 6x    (b) 10x    (c) 12x²
Start with the numerical factors, then combine with the variable factors
2
Find the HCF of each pair of terms:
(a) 4x and 6x    (b) 10x and 15    (c) 8x and 12x
3
Factorise each expression and verify by expanding:
(a) 4x + 8    (b) 6x − 9    (c) 10x + 15    (d) 3x + 12
4
The area of a rectangle is 5x + 10 cm². Factorise this expression to find a possible length and width of the rectangle.
🔵
Core — Include algebraic and combined HCFs. Show the HCF working separately before writing the bracket. Verify all answers.
1
Factorise fully (find the HCF including variable part):
(a) x² + 4x    (b) 3x² − 6x    (c) 5x² + 10x    (d) 8x² − 4x
2
Factorise fully and verify:
(a) 6x² + 9x
(b) 10x² − 15x
(c) 4x² + 8x − 12
(d) 6x² − 3x + 9
3
A student factorises 8x² + 12x as 2x(4x + 6). Explain why this is only a partial factorisation and write the fully factorised form.
4
Factorise then use the result to evaluate efficiently.
Factorise 7x + 7, then use the factorised form to quickly calculate the value when x = 13.
The factorised form should make mental arithmetic much easier — why?
5
Working backwards: an expression has been factorised as 3x(2x + 5). Write the original expanded expression and explain how you can confirm both forms are equivalent.
🔴
Extension — Multi-term expressions, algebraic factor expressions, partial vs full factorisation proofs, and connecting back to unit themes.
1
Factorise fully:
(a) 12x³ − 8x² + 4x
(b) 6a²b + 9ab²
(c) 5x(x + 2) + 3(x + 2)
For (c): treat (x + 2) as a common factor — it appears in both terms
2
In Lesson 6 Extension you showed that a(a+b) + b(a+b) simplifies to a² + 2ab + b². Now approach it from the factorising direction: what common factor do both terms of a(a+b) + b(a+b) share? Factor it out to show that the expression equals (a+b)(a+b) = (a+b)².
3
Prove using factorisation that the expression n² + n is always even for any positive integer n.
Factorise n² + n first. What do you notice about the resulting product?
4
Unit reflection: The seven lessons of this unit form a connected chain — each skill builds on the previous ones. Trace the connections: how does understanding pronumerals (L1) support substitution (L2)? How does substitution support pattern work (L3)? How do the properties in L4 underpin L5? How does L5 underpin expansion (L6)? And how does L6 underpin factorisation (L7)? Write a brief paragraph explaining the architecture of the unit.
🎉

Unit Complete!

You've worked through all seven lessons of Algebraic Techniques — from pronumerals and notation all the way to factorising with algebraic HCFs. That's the full MA4-ALG-C-01 outcome.

MA4-ALG-C-01 · Diagnostic

Unit Pre-Quiz

Seven questions — one for each lesson topic. This isn't a test: it shows you and your teacher what you already know so learning can be targeted. Work through every question carefully.

Q 0 / 7
MA4-ALG-C-01 · Assessment

Unit Post-Quiz

Seven new questions across all seven lesson topics — same skills, different problems. Show what you've learned since the pre-quiz.

Q 0 / 7