---
title: "Factorising"
subtitle: "Algebra Foundations · Brackets · Lesson 11"
---

# Factorising

**Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}}

> **Key question:** 6 x-tiles and 9 ones. Can you share them into equal rows, with nothing left over? The number of rows goes outside the bracket.


:::teacher-note

**In one sentence:** Students share algebra tiles into equal rows to factorise sx + t, learn to take out the highest common factor (numbers, then letters) and check by expanding (AC9M8A01, node 2).

**Students should be able to say:** I can factorise an expression by taking out a common factor. I can take out the highest common factor, including a letter, so the expression is fully factorised. I can check a factorisation by expanding it.

**Width:** the unmarked parts are the floor every student does (predict, run, investigate and make). Typical adds the modify practice; wider adds the real-world contexts. They add contexts, not difficulty. Students work in the online tutorial alongside this sheet.

:::


## Predict

Think about it: which number goes into both 6 and 9?

:::keep-together

**Which is $6x + 9$ written as a bracket, with the biggest possible number outside?**

☐ $3(2x + 3)$

☐ $3(2x + 9)$

☐ $2(3x + 4.5)$

☐ $6(x + 9)$

My prediction, and why:

{{lines-3}}

:::


:::teacher-note

**What happens:** 3(2x + 3). 3 goes into 6 and into 9: 6x ÷ 3 = 2x and 9 ÷ 3 = 3. Check by expanding: 3 × 2x = 6x and 3 × 3 = 9.

:::


## Run

Try it on the board in the tutorial. Was your prediction right? What was different?

{{lines-2}}

## The highest common factor


:::keep-together

### Take out the HCF

Here is how to factorise, then check.

1. $12x - 8$ — Find a number that goes into every term.
2. $\text{HCF of } 12 \text{ and } 8 = 4$ — Take out the biggest: the highest common factor.
3. $12x \div 4 = 3x$ — Divide every term by 4.
4. $-8 \div 4 = -2$ — The sign comes too.
5. $4(3x - 2)$ — The HCF goes outside; what is left goes inside.
6. $4 \times 3x - 4 \times 2 = 12x - 8$ — Check by expanding.

Try one: factorise $10x + 15$.

**Factorise $10x + 15$ fully.**

Answer: {{field-15}}

:::


:::teacher-note

**Answer:** 5(2x + 3)

The HCF of 10 and 15 is 5: 10x ÷ 5 = 2x and 15 ÷ 5 = 3. Check: 5 × 2x + 5 × 3 = 10x + 15.

:::


:::keep-together

### Find the mistake

**Ava factorises $12x - 8$. Click the line where her working goes wrong.**

```
 1  Factorise 12x − 8.
 2  The biggest number that goes into 12 and 8 is 2.
 3  12 ÷ 2 = 6 and 8 ÷ 2 = 4.
 4  So 12x − 8 = 2(6x − 4).
```

Line: {{field-15}}

Why is it wrong?

{{lines-2}}

:::


:::teacher-note

**Answer:** line 2

Line 2: 2 goes into both, but 4 is the biggest. 6x − 4 still shares a 2, so it is not fully factorised: 12x − 8 = 4(3x − 2).

:::


:::keep-together

### Check it

**How can you check that $4(3x - 2)$ is right for $12x - 8$?**

☐ Expand it and compare

☐ Check that 3 and 2 are both prime

☐ Add 4 to each term

:::


:::teacher-note

**Answer:** Expand it and compare

Expand: 4 × 3x = 12x and 4 × (−2) = −8. You are back to 12x − 8.

:::


:::typical

## One change at a time


**On the board.** Each task changes one thing.

**A minus sign.** No board from here: divide each term, then check by expanding.

**Three terms.** 

**A letter in common.** 

**Nothing to take out.** 

Work in the tutorial. Plan first: what will you change, and how will you test it?

{{box-4}}

:::


## Factorise and check


:::keep-together

### Positive

**Factorise $15x + 20$ fully.**

Answer: {{field-15}}

:::


:::teacher-note

**Answer:** 5(3x + 4)

HCF 5: 5(3x + 4).

:::


:::keep-together

### A minus sign

**Factorise $8x - 12$ fully.**

Answer: {{field-15}}

:::


:::teacher-note

**Answer:** 4(2x - 3)

HCF 4: 4(2x − 3).

:::


:::keep-together

### Full working

**Factorise $18x + 12$ fully. You may take out a factor in steps. Write one step per line.**

Answer: {{field-15}}

:::


:::teacher-note

**Answer:** 18x + 12 → 2(9x + 6) → 6(3x + 2)

2(9x + 6) still shares a 3, so take it out too: 6(3x + 2). Or take out the HCF 6 straight away.

:::


:::keep-together

### Review: expanding

**Review: expand $3(2x - 5)$.**

Answer: {{field-15}}

:::


:::teacher-note

**Answer:** 6x - 15

3 × 2x = 6x and 3 × (−5) = −15.

:::


## Make: four rows

Find an expression $sx + t$ that factorises fully as $4(\ldots)$. Then find another.

Stretch: why can't $s$ and $t$ both be multiples of 8?

Work in the tutorial. Plan first: what will you change, and how will you test it?

{{box-4}}

## Exit ticket

:::keep-together

**Factorise 14x − 21 fully, then check by expanding.**

{{lines-3}}

:::


:::teacher-note

**Answer:** 7(2x − 3). Check: 7 × 2x = 14x and 7 × (−3) = −21.

:::