Mathematics • Year 10 • Unit 2 • Lesson 4
Factorising Common Factors — Skill Drill
Build fluency with the reverse of expanding: pull the HCF out the front, leave the rest in a bracket. ax + ay = a(x + y). Check by expanding back. One worked example, one guided trace, eight independent problems.
1. I do — fully worked example
Find the HCF of every term, pull it out, leave the rest in a bracket. Always check by expanding.
Problem. Factorise 10a²b + 15ab².
Step 1 — Find the HCF of the numbers.
HCF(10, 15) = 5
Reason: 5 is the largest number that divides both 10 and 15.
Step 2 — Find the HCF of the letters: take the lowest power of each common letter.
a: lowest power is a¹. b: lowest power is b¹. Letter HCF = ab.
Reason: a² and a → both have at least a¹. b and b² → both have at least b¹.
Step 3 — Multiply the number and letter HCFs to get the overall HCF.
HCF = 5ab
Step 4 — Divide each term by the HCF to find what stays inside the bracket.
10a²b ÷ 5ab = 2a 15ab² ÷ 5ab = 3b
Step 5 — Write the factorised form, then check by expanding.
= 5ab(2a + 3b)
Check: 5ab(2a + 3b) = 10a²b + 15ab² ✓
Answer: 10a²b + 15ab² = 5ab(2a + 3b).
2. We do — fill in the missing steps
This time the HCF is a number only. Fill in each blank. 5 marks
Problem. Factorise 12x + 18.
Step 1 — HCF of the numbers. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. The largest factor in both lists is ____.
Step 2 — Any letters in common? 12x has the letter x; 18 has ____ letters. So no letters can come out.
Step 3 — Divide each term by the HCF:
12x ÷ ____ = ____ x
18 ÷ ____ = ____
Step 4 — Write the factorised form:
12x + 18 = ____ (____ x + ____)
Step 5 — Check by expanding (you should get 12x + 18 back):
Expansion check: __________________ = 12x + 18 ✓
3. You do — independent practice
Show your working. First four are foundation. Next two are standard (variables in the HCF). Last two are extension.
Foundation — numerical HCF only
3.1 Factorise 8x + 12. 1 mark
3.2 Factorise 6x + 9. 1 mark
3.3 Factorise 14a − 21. 1 mark
3.4 Factorise x² + 3x. (HCF includes a variable now.) 1 mark
Standard — number and letter HCF
3.5 Factorise 10a²b + 15ab. 2 marks
3.6 Factorise 6x²y − 9xy². 2 marks
Extension — push your thinking
3.7 Factorise by grouping: ax + ay + bx + by. (Hint: split into two pairs, factorise each pair, then pull a common bracket out.) 3 marks
3.8 A student writes 8x + 12 = 2(4x + 6). In one sentence, explain why this is not fully factorised, then write the fully factorised form. 2 marks
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Section 2 — We do (12x + 18)
Step 1: HCF = 6.
Step 2: 18 has no letters, so no letter HCF.
Step 3: 12x ÷ 6 = 2x; 18 ÷ 6 = 3.
Step 4: 12x + 18 = 6(2x + 3).
Step 5: Check: 6(2x + 3) = 12x + 18 ✓.
3.1 — 8x + 12
HCF = 4. 8x ÷ 4 = 2x; 12 ÷ 4 = 3. Answer: 4(2x + 3).
3.2 — 6x + 9
HCF = 3. 6x ÷ 3 = 2x; 9 ÷ 3 = 3. Answer: 3(2x + 3).
3.3 — 14a − 21
HCF = 7. 14a ÷ 7 = 2a; 21 ÷ 7 = 3. Answer: 7(2a − 3).
3.4 — x² + 3x
Both terms contain at least one x. HCF = x. x² ÷ x = x; 3x ÷ x = 3. Answer: x(x + 3).
3.5 — 10a²b + 15ab
Number HCF = 5. Letter HCF = ab (lowest power of each common letter). Total HCF = 5ab.
10a²b ÷ 5ab = 2a; 15ab ÷ 5ab = 3. Answer: 5ab(2a + 3).
3.6 — 6x²y − 9xy²
Number HCF = 3. Letter HCF = xy. Total HCF = 3xy.
6x²y ÷ 3xy = 2x; 9xy² ÷ 3xy = 3y. Answer: 3xy(2x − 3y).
3.7 — Factorise by grouping ax + ay + bx + by
Group into pairs: (ax + ay) + (bx + by).
Factor each pair: a(x + y) + b(x + y).
Now (x + y) is a common bracket — pull it out: (x + y)(a + b).
Check by expanding: (x + y)(a + b) = ax + bx + ay + by = ax + ay + bx + by ✓.
3.8 — Why 2(4x + 6) is not fully factorised
The inner terms 4x and 6 still share a common factor of 2 — so the bracket is not yet in simplest form. The HCF of 8x and 12 is 4, not 2.
Fully factorised form: 4(2x + 3).