Expanding & Factorising Brackets
Revise Expanding & Factorising Brackets for International Mathematics 0607 (O Level) — revision notes and instant AI marking. Free to start.
Expanding & Factorising Brackets
The big idea: Expanding removes brackets by multiplying everything out; factorising is the exact reverse — it puts an expression back into brackets. Once you can do both fluently, you can flip between the two forms whenever a problem needs it.
📋 Summary — What This Chapter Covers
- Expanding single brackets — multiplying one term by everything inside a bracket, e.g.
3x(x+2) - Expanding double brackets — using FOIL or a grid to multiply two brackets together
- Expanding triple brackets — expand two brackets first, then multiply the result by the third
- Factorising out terms — pulling the highest common factor (HCF) out to the front
- Factorising by grouping — splitting a 4-term expression into two pairs, each with its own common factor
- Factorising simple quadratics (where
a = 1) — finding two numbers that multiply tocand add tob - Factorising harder quadratics (where
a ≠ 1) — using the "multiplyac, add tob" trick plus grouping - Difference of two squares — spotting
a² − b²and instantly factorising to(a+b)(a−b) - Deciding which method to use — a decision process for any factorising question you're handed
1. Expanding Single Brackets
Think of the term outside the bracket as a "delivery van" — it has to visit every single term inside the bracket and multiply with it. Nothing inside gets left out.
3x(x + 2) means "3x multiplied by the whole bracket (x + 2)."
So 3x multiplies the x, AND 3x multiplies the 2.
Watch the signs!
This is where most marks are lost. Remember:
− × − = +(two negatives make a positive)− × + = −(a negative and a positive make a negative)
4x × (−3)
instead of 4x × −3. It stops sign errors creeping in when you're moving fast.
Expand −7x(4 − 5y)
Expand & Simplify (multiple brackets added together)
Sometimes you'll be given two separate bracket expressions added or subtracted, like
4(x + 7) + 5x(3 − x). The process is simple: expand each bracket completely
separately first, then collect like terms at the end.
Expand and simplify 3x(x + 2) − 7(x − 6)
2. Expanding Double Brackets
Now the "delivery van" idea doubles up: every term in the first bracket has to multiply every term in the second bracket. For two brackets each with 2 terms, that's always 4 multiplications in total.
Method 1: Using a grid
Write one bracket's terms as row headings and the other's as column headings. Multiply row × column in each cell, then add everything inside the grid together.
| x | +1 | |
|---|---|---|
| x | x² | x |
| +3 | 3x | 3 |
Method 2: FOIL
FOIL is just a memory trick for the same 4 multiplications, done in a fixed order:
Outside — multiply the outermost two terms
Inside — multiply the innermost two terms
Last — multiply the last terms in each bracket
Expand (2x − 3)(x + 4) using FOIL
xy terms
combine, giving 12x² − 10xy − 12y².
Squared brackets — a classic trap
(x + y)² is NOT the same as x² + y².
You must always write it out as two brackets first: (x + y)(x + y), then expand properly.
Expand (2x + 3)²
3. Expanding Triple Brackets
Three brackets multiplied together looks scary, but it breaks down into a task you already know how to do — twice.
2️⃣ You now have a single bracket (e.g. a quadratic) multiplied by the third, leftover bracket.
3️⃣ Multiply every term of that quadratic by every term of the remaining bracket (a grid is great here).
4️⃣ Collect like terms to get your final, simplified cubic expression.
Expand (2x − 3)(x + 4)(3x − 1)
| 3x | −1 | |
|---|---|---|
| 2x² | 6x³ | −2x² |
| 5x | 15x² | −5x |
| −12 | −36x | 12 |
4. Factorising Out Terms (Basic Factorising)
Factorising is expanding in reverse. Instead of removing brackets, you're putting an expression back into brackets by pulling out whatever is common to every term.
3(x + 2) is factorised — it's written as a product of factors.3x + 6 is not factorised — it's still an expanded sum.
Step-by-step method
- Find the highest common factor (HCF) of the numbers in each term.
- Find the highest common factor of the algebra letters (the lowest power of each letter that appears in every term).
- Multiply these together to get the overall HCF.
- Write the HCF outside a bracket, and divide each original term by the HCF to find what goes inside.
Factorise fully 5x⁴y + 25x³y³
x(6x + 10) is only partially
factorised — 6 and 10 still share a common factor of 2. The full answer is 2x(3x + 5).
Always double-check the bracket contents can't be broken down further!
5. Factorising by Grouping
This method is for expressions with 4 terms that don't all share one common factor, but split neatly into two pairs that each have their own common factor — and once you factorise each pair, they reveal a matching bracket that can be pulled out too.
(t + 4), so it can be
"taken out" just like a common factor: (t + 4)(3x + 2)
Step-by-step method
- Look at the 4 terms and find a common factor between the first two.
- Find a common factor between the last two.
- Factorise each pair separately — you should now see the same bracket in both.
- Factorise that matching bracket out to finish.
Factorise ab + 3b + 2a + 6
6. Factorising Simple Quadratics (a = 1)
A quadratic in the form x² + bx + c factorises into two brackets:
(x + p)(x + q). Your job is to find the mystery numbers p
and q.
Factorise x² − 4x − 21
Alternative: factorise by grouping
If inspection feels tricky, split the middle term using your two numbers, then group in pairs — exactly like Topic 5.
Factorise x² − 5x + 6
7. Factorising Harder Quadratics (a ≠ 1)
When the quadratic starts with a coefficient other than 1, e.g. 4x² − 25x − 21,
the "multiply/add" trick still works — but you multiply a × c instead of just c.
Factorise 6x² − 7x − 3
Or use a grid instead of grouping
Split the middle term the same way, but arrange the four terms in a 2×2 grid (like reversing the expanding grid) and read the row/column headings off as your two brackets.
8. Difference of Two Squares
This is a shortcut pattern worth memorising cold — when you spot it, you can factorise (or expand) almost instantly without any of the "find two numbers" work above.
Proof it works — just expand the right-hand side:
Factorise 9x² − 16
2y² − 50 doesn't look like a difference of two squares at first glance.
But take out the common factor of 2 first: 2(y² − 25) — now
y² − 25 = y² − 5² is clearly recognisable, giving
2(y + 5)(y − 5).
(a + b)(a − b) expand instantly to a² − b²
— no need for full FOIL. E.g. (2x + 3)(2x − 3) = 4x² − 9 in one line.
9. Deciding Which Factorisation Method to Use
In the exam you won't be told which method to use — you have to recognise the type of expression yourself. Here's the decision process, step by step:
→ Common factor visible? → Factorise out the HCF (Topic 4)
→ Looks like a² − b² (subtraction of two squares)? → Difference of two squares (Topic 8)
3 terms (a quadratic)?
→ Starts with x² (a = 1)? → Factorise simple quadratic (Topic 6)
→ Starts with ax² (a ≠ 1)? → Check for a common factor first, THEN factorise harder quadratic (Topic 7)
4 terms?
→ Factorise by grouping (Topic 5)
A cubed term (x³) present?
→ Factorise out the common factor first (often just x), then factorise the remaining quadratic
How do I know FOR SURE if it will factorise?
For ax² + bx + c, check the discriminant:
Factorise −8x² + 100x − 48
2x² + 4x + 2 does NOT become
x² + 2x + 1 unless you factorise out the 2 properly and KEEP it:
2(x² + 2x + 1).
🧠 What To Memorise
| Term / Rule | Meaning |
|---|---|
| Expand | Multiply out brackets to remove them, producing a sum of terms |
| Factorise | The reverse of expanding — write an expression as a product of factors (brackets) |
| Highest Common Factor (HCF) | The biggest factor shared by every term — numbers AND letters |
| a(b + c) = ab + ac | Single bracket expansion rule |
| FOIL | First, Outside, Inside, Last — order for expanding two brackets |
| Simple quadratic (a=1) | Find 2 numbers: multiply → c, add → b |
| Harder quadratic (a≠1) | Find 2 numbers: multiply → a×c, add → b, then split & group |
| a² − b² = (a+b)(a−b) | Difference of two squares — subtraction of perfect squares only |
| Discriminant: b² − 4ac | If this is a perfect square, the quadratic factorises into integers |
| (x + a)² factorised form | x² + 2ax + a² = (x + a)² — spot this pattern to save time |
✅ Concepts Checklist
Tick off each concept once you're confident you could explain it to someone else, no notes needed.
🎯 Exam Tips & Common Mistakes
x(6x + 10) will lose you
marks — the correct fully factorised answer is 2x(3x + 5). Always check whether
the bracket contents share a further common factor.
− × − = + and − × + = −. This single habit prevents
most of the errors students make in this topic.
a² − b² = (a+b)(a−b) works ONLY for subtraction. There is no equivalent
factorisation for a² + b² — it simply cannot be factorised over normal numbers.
Don't try to force it!
- "Expand and simplify..." (single or double brackets, sometimes with two expressions added)
- "Factorise..." (2, 3, or 4-term expressions — you must identify the right method yourself)
- "Factorise fully..." (checking you spot ALL common factors, not just some)
- Multi-step problems combining factorising with solving equations or simplifying fractions
- Difference of two squares disguised with a common factor, e.g. 2y² − 50
- 🎯 Exam Tips & Common Mistakes
- Expand & Simplify (multiple brackets added together)
Read the full Expanding & Factorising Brackets notes free
That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.
Unlock the full notes free →