Chapter 3 · Expansion & Factorisation

1.1.7 Expand, Then Reverse It

Multiplying two brackets out and splitting a trinomial back into two brackets are the same journey, walked in opposite directions.

🎯 By the end of this you can…

✓Expand a binomial times a trinomial by distributing every term
✓Factorise a quadratic trinomial by splitting the middle term
✓Spot which pair of numbers is a trap — right product, wrong sum

💡 The idea

Two directions of the same road

Expanding $(x + 2) (x^{2} + x + 1)$ multiplies every term in the first bracket by every term in the second, then collects like terms. Factorising walks that exact road backwards: start from the collected trinomial and rebuild the two brackets.

✏️ Worked example

Expand (x + 2)(x² + x + 1)

1
$x (x^{2} + x + 1) = x^{3} + x^{2} + x$
2
$2 (x^{2} + x + 1) = 2 x^{2} + 2 x + 2$
3
Add and collect like terms: $x^{3} + (x^{2} + 2 x^{2}) + (x + 2 x) + 2$

✅ Answer: $x^{3} + 3 x^{2} + 3 x + 2$

📋 Factorising by splitting the middle term

1
Multiply the FIRST coefficient by the LAST coefficient (counting signs).
2
Find two numbers that multiply to that product AND add to the MIDDLE coefficient.
3
Split the middle term into those two numbers' terms.
4
Group the four terms in pairs and pull out a common factor from each pair — the two pairs should leave a matching bracket behind.

🔍 Factorising $a^{2} + 3 ab - 4 b^{2}$ — searching for the split

First × last: $1 \times - 4 = - 4$ . Hunt for two numbers that multiply to $- 4$ AND add to $3$ (the middle coefficient).

1 and -4product

1 \times - 4 = - 4

✅sum

1 + - 4 = - 3

❌ (need 3)

tried 1 / 3

✏️ Worked example

Finish factorising a² + 3ab − 4b², now that the split is found (4 and −1)

1
Split the middle term: $a^{2} + 3 ab - 4 b^{2} = a^{2} + 4 ab - ab - 4 b^{2}$ .
2
Group in pairs: $(a^{2} + 4 ab) + (- ab - 4 b^{2}) = a (a + 4 b) - b (a + 4 b)$ .
3
Both pairs share the bracket $(a + 4 b)$ — pull it out.

✅ Answer: $(a - b) (a + 4 b)$

⚠️ The classic trap

For $a^{2} + 3 ab - 4 b^{2}$ , the pair $- 2$ and $2$ also multiplies to $- 4$ — but they add to $0$ , not $3$ . A pair only works if it satisfies BOTH conditions at once: right product AND right sum.

✏️ Worked example

Factorise 3p² + 15pq + 18q²

1
First, pull out the common factor $3$ : $3 (p^{2} + 5 pq + 6 q^{2})$ .
2
Inside the bracket: numbers that multiply to $6$ and add to $5$ are $2$ and $3$ .
3
$p^{2} + 2 pq + 3 pq + 6 q^{2} = p (p + 2 q) + 3 q (p + 2 q) = (p + 2 q) (p + 3 q)$

✅ Answer: $3 (p + 2 q) (p + 3 q)$

🎮 Expand it, or factorise it

⭐ 0 XP🔥 streak 0

Expand

(x + 9 y) (x + 3 y + 1)

Factorise

c^{2} - 4 c d - 21 d^{2}

Factorise

2 h^{2} + 7 hk - 15 k^{2}

Factorise

x^{2} + 5 x + 6

Factorise

x^{2} - x - 6

0 of 5 finished

🤔 To factorise a² + 3ab − 4b², which pair of numbers correctly splits the middle term?

🏅

You finished this lesson!

Expanding and factorising are now two ends of the same skill — see a trinomial, and you can walk straight back to the two brackets it came from.

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