Welcome to Expanding and Factorising!
Welcome to one of the most exciting and powerful areas of algebra! If you have ever packed a lunchbox into a backpack or unpacked it at lunchtime, you already understand the core idea behind this chapter. In algebra, expanding is like unpacking everything so that every part is out in the open, while factorising is like neatly packing everything back into boxes (brackets).
Don't worry if algebra has felt confusing in the past. We will take this step by step, using visual methods, memory tricks, and clear examples. By the end of these notes, you will be able to multiply out double brackets and factorise quadratic expressions with complete confidence!
1. Key Words and Building Blocks
Before we jump into calculations, let's look at the mathematical words you will see throughout this topic:
Term: A single mathematical building block. It can be an individual number, a single variable (letter), or numbers and letters multiplied together. For example: \(3x\), \(7\), and \(-2y^2\) are all terms.
Binomial: An algebraic expression that contains exactly two distinct terms joined by a plus or minus sign. For example: \((x + 3)\) or \((2x - 5)\).
Expanding (Multiplying Out): The process of removing brackets by multiplying every term inside by what is on the outside, using the distributive law.
Factorising: The exact opposite (inverse) of expanding. It means putting an expression back into brackets as a product of its factors.
Fully Factorised: When you have taken out the Highest Common Factor (HCF) so that nothing more can be divided out.
Quadratic Expression: An expression where the highest power of the variable is \(2\). Its standard algebraic form is \(ax^2 + bx + c\) (where \(a \neq 0\)).
Key Takeaway: Expanding and factorising are inverse operations. Expanding removes brackets; factorising brings brackets back!
2. Quick Warm-Up: Expanding Single Brackets
To expand a single bracket, we use the distributive law: multiply the term outside the bracket by every single term inside the bracket.
\(a(b + c) = ab + ac\)
\(a(b - c) = ab - ac\)
Example 1: Basic Single Bracket
Expand \(3(2x + 5)\):
Multiply \(3 \times 2x = 6x\)
Multiply \(3 \times 5 = 15\)
Result: \(6x + 15\)
Watch Out for Negative Multipliers!
When there is a negative number outside, remember that multiplying two negative numbers gives a positive result:
\(-a(b - c) = -ab + ac\)
Expand \(-2(3x - 4)\):
Multiply \(-2 \times 3x = -6x\)
Multiply \(-2 \times -4 = +8\)
Result: \(-6x + 8\)
3. Expanding Double Brackets (Product of Two Binomials)
When two binomial brackets are next to each other, like \((x + 2)(x + 3)\), it means they are multiplied together. The golden rule is: every term in the first bracket must multiply every term in the second bracket.
There are three standard ways to do this at Key Stage 3. Choose the method that works best for you!
Method 1: The Grid / Area Model
Think of the brackets as the width and length of a rectangle split into four smaller rooms:
Let's expand \((x + 4)(x + 3)\):
Set up a \(2 \times 2\) grid with \(x\) and \(+4\) across the top, and \(x\) and \(+3\) down the side:
Top-left cell: \(x \times x = x^2\)
Top-right cell: \(x \times 4 = +4x\)
Bottom-left cell: \(3 \times x = +3x\)
Bottom-right cell: \(3 \times 4 = +12\)
Now write down all four terms and collect the like terms in the middle:
\(x^2 + 4x + 3x + 12 = x^2 + 7x + 12\)
Method 2: The FOIL Method
FOIL is a handy memory trick that helps you remember the four multiplications in order:
F - First terms in each bracket
O - Outside terms (the two outer edges)
I - Inside terms (the two inner terms)
L - Last terms in each bracket
Let's expand \((x + 5)(x - 2)\) using FOIL:
F: \(x \times x = x^2\)
O: \(x \times (-2) = -2x\)
I: \(5 \times x = +5x\)
L: \(5 \times (-2) = -10\)
Combine them together: \(x^2 - 2x + 5x - 10\)
Simplify the linear \(x\) terms: \(-2x + 5x = +3x\)
Final Answer: \(x^2 + 3x - 10\)
Method 3: Distributive Splitting
You can also split the first bracket into two single-bracket problems:
\((x + a)(x + b) = x(x + b) + a(x + b)\)
For example, \((x + 6)(x - 3) = x(x - 3) + 6(x - 3)\)
\(= x^2 - 3x + 6x - 18\)
\(= x^2 + 3x - 18\)
4. Special Cases in Expanding
Special Case A: Squaring a Binomial
When you see \((x + a)^2\), it means the entire bracket is multiplied by itself: \((x + a)(x + a)\).
\((x + a)^2 = (x + a)(x + a) = x^2 + 2ax + a^2\)
\((x - a)^2 = (x - a)(x - a) = x^2 - 2ax + a^2\)
Example: Expand \((x + 5)^2\)
Write it out fully: \((x + 5)(x + 5)\)
Expand: \(x^2 + 5x + 5x + 25 = x^2 + 10x + 25\)
Special Case B: Difference of Two Squares (DOTS)
When the two brackets are identical except for one having a \(+\) and the other a \(-\), the middle terms cancel each other out completely!
\((x + a)(x - a) = x^2 - ax + ax - a^2 = x^2 - a^2\)
Example: Expand \((x + 7)(x - 7)\)
FOIL: \(x^2 - 7x + 7x - 49\)
Notice that \(-7x + 7x = 0\), leaving: \(x^2 - 49\)
Key Takeaway: When expanding double brackets, you always get four terms first, which usually simplify down to three terms (a quadratic: \(x^2 + bx + c\)).
5. Factorising into a Single Bracket (Taking Out the HCF)
To factorise an expression into a single bracket, we look for the Highest Common Factor (HCF) of both the numbers and the letters.
Step-by-Step Guide:
Step 1: Find the highest number that divides into all coefficients.
Step 2: Find any common letters shared by all terms.
Step 3: Place this full HCF outside the bracket.
Step 4: Divide each original term by the HCF to find what stays inside.
Example 1: Numbers and Letters
Factorise \(6x^2 + 9x\)
1. HCF of \(6\) and \(9\) is \(3\).
2. Both terms share an \(x\) (since \(x^2 = x \times x\)). So the complete HCF is \(3x\).
3. Put \(3x\) outside: \(3x( \quad + \quad )\)
4. Divide: \(6x^2 \div 3x = 2x\) and \(9x \div 3x = 3\)
Final Answer: \(3x(2x + 3)\)
Example 2: When a Term is the Factor
Factorise \(4x + 4\)
The HCF is \(4\).
Divide \(4x \div 4 = x\)
Divide \(4 \div 4 = 1\)
Final Answer: \(4(x + 1)\) (Never leave the space empty when a term divides completely!)
6. Factorising Monic Quadratics (\(x^2 + bx + c\))
A monic quadratic is a quadratic where the coefficient of \(x^2\) is \(1\). We want to factorise \(x^2 + bx + c\) into two brackets: \((x + p)(x + q)\).
The Golden Rule for Quadratic Factorising:
Find two numbers \(p\) and \(q\) that satisfy two conditions:
1. Multiply to give the last number: \(p \times q = c\)
2. Add to give the middle number: \(p + q = b\)
Sign Rules Helper:
Case 1: Both signs positive (\(x^2 + bx + c\))
Both numbers are positive: \((x + \text{positive})(x + \text{positive})\)
Example: Factorise \(x^2 + 7x + 12\)
Find pairs that multiply to \(12\): \(1 \times 12\), \(2 \times 6\), \(3 \times 4\).
Which pair adds to \(7\)? \(3 + 4 = 7\).
Answer: \((x + 3)(x + 4)\)
Case 2: Middle negative, end positive (\(x^2 - bx + c\))
Both numbers must be negative (since negative \(\times\) negative \(=\) positive, but negative \(+\) negative \(=\) negative): \((x - \text{number})(x - \text{number})\)
Example: Factorise \(x^2 - 8x + 15\)
Factors of \(+15\) that add to \(-8\): \(-3\) and \(-5\) (since \(-3 \times -5 = +15\) and \(-3 + -5 = -8\)).
Answer: \((x - 3)(x - 5)\)
Case 3: End term is negative (\(x^2 + bx - c\) or \(x^2 - bx - c\))
One number must be positive and one must be negative.
Example: Factorise \(x^2 + 2x - 15\)
Factors of \(-15\): \(5 \times (-3) = -15\) and \(5 + (-3) = +2\).
Answer: \((x + 5)(x - 3)\)
Factorising Difference of Two Squares
If you see a quadratic with no middle \(x\) term, where a square number is subtracted from \(x^2\), use the Difference of Two Squares rule:
\(x^2 - a^2 = (x + a)(x - a)\)
Example: Factorise \(x^2 - 36\)
Since \(36 = 6^2\), this factorises directly to: \((x + 6)(x - 6)\)
Key Takeaway: Always look at the constant term \(c\) first. Find pairs of numbers that multiply to make \(c\), then pick the pair that adds to give \(b\).
7. Common Mistakes and How to Avoid Them
Mistake 1: The "Freshman's Square" Error
Incorrect: \((x + 5)^2 = x^2 + 25\)
Why it's wrong: Squaring a bracket means multiplying it by itself! You must not forget the middle term.
Correct: \((x + 5)^2 = (x + 5)(x + 5) = x^2 + 10x + 25\)
Mistake 2: Mixing Up Negative Signs in Double Brackets
Incorrect: \((x - 3)(x - 4) = x^2 - 7x - 12\)
Why it's wrong: The last step is \(-3 \times -4\), which equals \(+12\), not \(-12\).
Correct: \(x^2 - 7x + 12\)
Mistake 3: Partial Factorisation
Incorrect: Factorising \(12x + 18\) as \(2(6x + 9)\) or \(3(4x + 6)\)
Why it's wrong: While mathematically equivalent, it is not fully factorised because \(6\) and \(9\) still share a common factor of \(3\).
Correct: Take out the full HCF of \(6\): \(6(2x + 3)\)
Mistake 4: Swapping the Sum and Product Rules
Remember: You multiply to get the end number (\(c\)) and add to get the middle number (\(b\)). A quick rhyme to remember: "Multiply to the back, add to the middle track!"
Mistake 5: Forgetting the Number 1
Incorrect: \(4x + 4 = 4(x)\)
Correct: \(4x + 4 = 4(x + 1)\), because \(4 \div 4 = 1\).
Quick Review Summary
Expanding Single Brackets: Multiply outside term by every inside term: \(a(b + c) = ab + ac\).
Expanding Double Brackets: Use FOIL or the Grid Method to get four terms, then collect like terms in the middle: \((x + a)(x + b) = x^2 + (a + b)x + ab\).
Squaring a Binomial: \((x + a)^2 = (x + a)(x + a) = x^2 + 2ax + a^2\).
Difference of Two Squares: \((x + a)(x - a) = x^2 - a^2\).
Factorising into Single Bracket: Find the HCF of numbers and letters, place outside, divide inside.
Factorising Quadratics: For \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\) to write as \((x + p)(x + q)\).