3.2 Brackets and Factorising
What is Expanding?
Expanding brackets means multiplying out the brackets to remove them. You multiply everything inside the bracket by the term outside.
Expanding brackets means multiplying out the brackets to remove them. You multiply everything inside the bracket by the term outside.
⚡ The Rule:
$$a(b + c) = ab + ac$$
Multiply the term outside by each term inside the bracket.
Multiply the term outside by each term inside the bracket.
Example 1: Expand $3(x + 4)$
Step 1: Multiply 3 by the first term: $3 \times x = 3x$
Step 2: Multiply 3 by the second term: $3 \times 4 = 12$
Answer: $3(x + 4) = 3x + 12$
Step 2: Multiply 3 by the second term: $3 \times 4 = 12$
Answer: $3(x + 4) = 3x + 12$
Example 2: Expand $4x(2x + 3)$
Step 1: Multiply $4x$ by $2x$: $4x \times 2x = 8x^2$
Step 2: Multiply $4x$ by $3$: $4x \times 3 = 12x$
Answer: $4x(2x + 3) = 8x^2 + 12x$
Step 2: Multiply $4x$ by $3$: $4x \times 3 = 12x$
Answer: $4x(2x + 3) = 8x^2 + 12x$
Example 3: Expand $-2(3x - 5)$
Step 1: Multiply $-2$ by $3x$: $-2 \times 3x = -6x$
Step 2: Multiply $-2$ by $-5$: $-2 \times (-5) = +10$
Answer: $-2(3x - 5) = -6x + 10$
Remember: negative × negative = positive!
Step 2: Multiply $-2$ by $-5$: $-2 \times (-5) = +10$
Answer: $-2(3x - 5) = -6x + 10$
Remember: negative × negative = positive!
💡 More Examples:
$$5(2x + 1) = 10x + 5$$
$$x(x + 7) = x^2 + 7x$$
$$-3(2x - 4) = -6x + 12$$
$$2x(3x - 2y + 5) = 6x^2 - 4xy + 10x$$
🎯 Expansion Practice
Real Life Uses:
• Area calculations: Finding the area of combined shapes
• Pricing: Calculating totals with discounts or taxes
• Distribution: Sharing items among groups
• Area calculations: Finding the area of combined shapes
• Pricing: Calculating totals with discounts or taxes
• Distribution: Sharing items among groups
The FOIL Method
When expanding two brackets, use FOIL to remember which terms to multiply:
When expanding two brackets, use FOIL to remember which terms to multiply:
⚡ FOIL:
First terms
Outer terms
Inner terms
Last terms
$$(a + b)(c + d) = ac + ad + bc + bd$$
Outer terms
Inner terms
Last terms
$$(a + b)(c + d) = ac + ad + bc + bd$$
Example 1: Expand $(x + 2)(x + 3)$
First: $x \times x = x^2$
Outer: $x \times 3 = 3x$
Inner: $2 \times x = 2x$
Last: $2 \times 3 = 6$
Combine: $x^2 + 3x + 2x + 6 = x^2 + 5x + 6$
Outer: $x \times 3 = 3x$
Inner: $2 \times x = 2x$
Last: $2 \times 3 = 6$
Combine: $x^2 + 3x + 2x + 6 = x^2 + 5x + 6$
Example 2: Expand $(x - 4)(x + 5)$
First: $x \times x = x^2$
Outer: $x \times 5 = 5x$
Inner: $-4 \times x = -4x$
Last: $-4 \times 5 = -20$
Combine: $x^2 + 5x - 4x - 20 = x^2 + x - 20$
Outer: $x \times 5 = 5x$
Inner: $-4 \times x = -4x$
Last: $-4 \times 5 = -20$
Combine: $x^2 + 5x - 4x - 20 = x^2 + x - 20$
Example 3: Expand $(2x + 3)(x - 4)$
First: $2x \times x = 2x^2$
Outer: $2x \times (-4) = -8x$
Inner: $3 \times x = 3x$
Last: $3 \times (-4) = -12$
Combine: $2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12$
Outer: $2x \times (-4) = -8x$
Inner: $3 \times x = 3x$
Last: $3 \times (-4) = -12$
Combine: $2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12$
💡 Special Case: Squaring a Bracket
$(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2$
$(a - b)^2 = (a - b)(a - b) = a^2 - 2ab + b^2$
Example: $(x + 3)^2 = x^2 + 6x + 9$
Example: $(x - 5)^2 = x^2 - 10x + 25$
⚠️ Common mistake: $(x + 3)^2 \neq x^2 + 9$
$(a - b)^2 = (a - b)(a - b) = a^2 - 2ab + b^2$
Example: $(x + 3)^2 = x^2 + 6x + 9$
Example: $(x - 5)^2 = x^2 - 10x + 25$
⚠️ Common mistake: $(x + 3)^2 \neq x^2 + 9$
🎯 Expansion Practice
Real Life Uses:
• Area: Finding area when dimensions are expressions
• Physics: Kinematic equations, projectile motion
• Finance: Compound interest calculations
• Area: Finding area when dimensions are expressions
• Physics: Kinematic equations, projectile motion
• Finance: Compound interest calculations
Strategy for Triple Brackets
For three brackets, expand two of them first, then multiply the result by the third bracket.
For three brackets, expand two of them first, then multiply the result by the third bracket.
⚡ The Method:
$(a)(b)(c)$
Step 1: Expand $(a)(b)$ first
Step 2: Multiply the result by $(c)$
Tip: Choose the easiest pair to expand first!
Step 1: Expand $(a)(b)$ first
Step 2: Multiply the result by $(c)$
Tip: Choose the easiest pair to expand first!
Example: Expand $(x + 1)(x + 2)(x + 3)$
Step 1: Expand $(x + 1)(x + 2)$ first
$= x^2 + 2x + x + 2$
$= x^2 + 3x + 2$
Step 2: Now multiply by $(x + 3)$
$= (x^2 + 3x + 2)(x + 3)$
Step 3: Expand each term
$= x^2(x + 3) + 3x(x + 3) + 2(x + 3)$
$= x^3 + 3x^2 + 3x^2 + 9x + 2x + 6$
Step 4: Collect like terms
$= x^3 + 6x^2 + 11x + 6$
$= x^2 + 2x + x + 2$
$= x^2 + 3x + 2$
Step 2: Now multiply by $(x + 3)$
$= (x^2 + 3x + 2)(x + 3)$
Step 3: Expand each term
$= x^2(x + 3) + 3x(x + 3) + 2(x + 3)$
$= x^3 + 3x^2 + 3x^2 + 9x + 2x + 6$
Step 4: Collect like terms
$= x^3 + 6x^2 + 11x + 6$
Example: Expand $(x - 1)(x + 1)(x + 2)$
Step 1: Notice $(x - 1)(x + 1)$ is a difference of two squares!
$= x^2 - 1$
Step 2: Now multiply by $(x + 2)$
$= (x^2 - 1)(x + 2)$
$= x^3 + 2x^2 - x - 2$
💡 Look for shortcuts like difference of two squares!
$= x^2 - 1$
Step 2: Now multiply by $(x + 2)$
$= (x^2 - 1)(x + 2)$
$= x^3 + 2x^2 - x - 2$
💡 Look for shortcuts like difference of two squares!
Real Life Uses:
• Volume: Calculating volumes of 3D shapes with variable dimensions
• Engineering: Complex mechanical calculations
• Economics: Multi-variable models
• Volume: Calculating volumes of 3D shapes with variable dimensions
• Engineering: Complex mechanical calculations
• Economics: Multi-variable models
What is Factorising?
Factorising is the opposite of expanding — you put an expression back into brackets by finding common factors.
Factorising is the opposite of expanding — you put an expression back into brackets by finding common factors.
⚡ The Method:
Step 1: Find the highest common factor (HCF) of all terms
Step 2: Write the HCF outside the bracket
Step 3: Divide each term by the HCF and write inside the bracket
Step 4: Check by expanding!
Step 2: Write the HCF outside the bracket
Step 3: Divide each term by the HCF and write inside the bracket
Step 4: Check by expanding!
Example 1: Factorise $6x + 9$
Step 1: Find the HCF of 6 and 9
Factors of 6: 1, 2, 3, 6
Factors of 9: 1, 3, 9
HCF = 3
Step 2: Write 3 outside the bracket
$3(...)$
Step 3: Divide each term by 3
$6x \div 3 = 2x$
$9 \div 3 = 3$
Answer: $6x + 9 = 3(2x + 3)$
Check: $3(2x + 3) = 6x + 9$ ✓
Factors of 6: 1, 2, 3, 6
Factors of 9: 1, 3, 9
HCF = 3
Step 2: Write 3 outside the bracket
$3(...)$
Step 3: Divide each term by 3
$6x \div 3 = 2x$
$9 \div 3 = 3$
Answer: $6x + 9 = 3(2x + 3)$
Check: $3(2x + 3) = 6x + 9$ ✓
Example 2: Factorise $12x^2 - 8x$
Step 1: Find the HCF of $12x^2$ and $8x$
• Numbers: HCF of 12 and 8 = 4
• Letters: Both have at least one $x$
• HCF = $4x$
Step 2: Divide each term by $4x$
$12x^2 \div 4x = 3x$
$8x \div 4x = 2$
Answer: $12x^2 - 8x = 4x(3x - 2)$
• Numbers: HCF of 12 and 8 = 4
• Letters: Both have at least one $x$
• HCF = $4x$
Step 2: Divide each term by $4x$
$12x^2 \div 4x = 3x$
$8x \div 4x = 2$
Answer: $12x^2 - 8x = 4x(3x - 2)$
Example 3: Factorise $15x^2y + 10xy^2 - 5xy$
Step 1: Find the HCF
• Numbers: HCF of 15, 10, 5 = 5
• Letters: All have $x$ and $y$
• HCF = $5xy$
Step 2: Divide each term by $5xy$
$15x^2y \div 5xy = 3x$
$10xy^2 \div 5xy = 2y$
$5xy \div 5xy = 1$
Answer: $15x^2y + 10xy^2 - 5xy = 5xy(3x + 2y - 1)$
• Numbers: HCF of 15, 10, 5 = 5
• Letters: All have $x$ and $y$
• HCF = $5xy$
Step 2: Divide each term by $5xy$
$15x^2y \div 5xy = 3x$
$10xy^2 \div 5xy = 2y$
$5xy \div 5xy = 1$
Answer: $15x^2y + 10xy^2 - 5xy = 5xy(3x + 2y - 1)$
🎯 Fctorising Brackets
What is DOTS?
The Difference of Two Squares is a special factorisation pattern that occurs when you have one squared term subtracted from another.
The Difference of Two Squares is a special factorisation pattern that occurs when you have one squared term subtracted from another.
⚡ The Formula:
$$a^2 - b^2 = (a + b)(a - b)$$
This ONLY works for subtraction, not addition!
This ONLY works for subtraction, not addition!
💡 How to Recognise DOTS:
✓ Two terms only
✓ Both terms are perfect squares
✓ One term is being subtracted from the other
Perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100...$
Also: $x^2, 4x^2, 9x^2, y^4, z^6...$ (even powers)
✓ Both terms are perfect squares
✓ One term is being subtracted from the other
Perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100...$
Also: $x^2, 4x^2, 9x^2, y^4, z^6...$ (even powers)
Example 1: Factorise $x^2 - 9$
Step 1: Identify the squares
• $x^2 = (x)^2$ ✓
• $9 = (3)^2$ ✓
Step 2: Apply the formula
$$x^2 - 9 = (x + 3)(x - 3)$$
Check: $(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9$ ✓
• $x^2 = (x)^2$ ✓
• $9 = (3)^2$ ✓
Step 2: Apply the formula
$$x^2 - 9 = (x + 3)(x - 3)$$
Check: $(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9$ ✓
Example 2: Factorise $25x^2 - 16$
Step 1: Identify the squares
• $25x^2 = (5x)^2$ ✓
• $16 = (4)^2$ ✓
Step 2: Apply the formula
$$25x^2 - 16 = (5x + 4)(5x - 4)$$
• $25x^2 = (5x)^2$ ✓
• $16 = (4)^2$ ✓
Step 2: Apply the formula
$$25x^2 - 16 = (5x + 4)(5x - 4)$$
Example 3: Factorise $4x^2 - 49y^2$
Step 1: Identify the squares
• $4x^2 = (2x)^2$ ✓
• $49y^2 = (7y)^2$ ✓
Step 2: Apply the formula
$$4x^2 - 49y^2 = (2x + 7y)(2x - 7y)$$
• $4x^2 = (2x)^2$ ✓
• $49y^2 = (7y)^2$ ✓
Step 2: Apply the formula
$$4x^2 - 49y^2 = (2x + 7y)(2x - 7y)$$
Example 4: Factorise $x^4 - 1$
Step 1: First application of DOTS
• $x^4 = (x^2)^2$ and $1 = (1)^2$
$$x^4 - 1 = (x^2 + 1)(x^2 - 1)$$
Step 2: Notice $(x^2 - 1)$ is also DOTS!
$$x^2 - 1 = (x + 1)(x - 1)$$
Final answer: $x^4 - 1 = (x^2 + 1)(x + 1)(x - 1)$
• $x^4 = (x^2)^2$ and $1 = (1)^2$
$$x^4 - 1 = (x^2 + 1)(x^2 - 1)$$
Step 2: Notice $(x^2 - 1)$ is also DOTS!
$$x^2 - 1 = (x + 1)(x - 1)$$
Final answer: $x^4 - 1 = (x^2 + 1)(x + 1)(x - 1)$
🎯 Is it DOTS?
🎯 Factorising with DOTS Factorise using DOTS