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Common Factors and Grouping

Maths · Year 9
Year 9
Lesson Guide
Explanation
Factorising is expanding backwards: you put the brackets back in. For twelve x plus eighteen, find the highest number that divides both terms. Twelve and eighteen share the factors one, two, three and six, and the highest is six. Divide each term by six: twelve x becomes two x and eighteen becomes three. So twelve x plus eighteen is six times the bracket two x plus three. Check by expanding it back.
Your Turn
Factorise 15y + 20. Check: 5(3y + 4).
Watch Out
12x + 18 = 2(6x + 9) is not finished: 6x + 9 still has a common factor of 3. Take out the highest common factor.
Explanation
Letters can be part of the common factor too. Ten p cubed minus fifteen p squared. The numbers ten and fifteen have a highest common factor of five. Both terms contain p, and the lowest power of p is p squared, so that is the most p you can take out. The HCF is five p squared. Divide each term by it: ten p cubed becomes two p, and negative fifteen p squared becomes negative three. The answer is five p squared times the bracket two p minus three.
Your Turn
Factorise 4k2 + 6k. Check: 2k(2k + 3).
Watch Out
Take the lowest power: 10p3 − 15p2 = 5p(2p2 − 3p) is true but not fully factorised.
Explanation
When the first term is negative, it is usual to take out a negative common factor. Negative six a squared b plus nine a b squared. Six and nine share three, and both terms have at least one a and one b, so the HCF is three a b. Take out negative three a b. Negative six a squared b divided by negative three a b is positive two a. Positive nine a b squared divided by negative three a b is negative three b. The answer is negative three a b times the bracket two a minus three b. Notice that every sign inside has flipped.
Your Turn
Factorise −4x − 10 by taking out a negative. Check: −2(2x + 5).
Watch Out
After taking out a negative, every sign inside flips. −6a2b + 9ab2 is not −3ab(2a + 3b).
Explanation
When there are four terms and nothing is common to all of them, group them in pairs. Three m plus three n plus k m plus k n. The first pair has a common factor of three: three times m plus n. The second pair has a common factor of k: k times m plus n. Now both parts contain the same bracket, m plus n, so that bracket is a common factor. Take it out, and what is left is three plus k. The answer is m plus n, times three plus k.
Your Turn
Factorise xy + 2x + 5y + 10. Check: (y + 2)(x + 5).
Watch Out
The brackets must match exactly before you can take one out. If they do not, try pairing the terms differently.
Explanation
Grouping has one trap. In p q minus four p minus five q plus twenty, the first pair gives p times q minus four. For the second pair, the first term is negative five q, so take out negative five. Negative five q divided by negative five is q, and twenty divided by negative five is negative four, giving negative five times q minus four. The brackets match, so the answer is q minus four, times p minus five. If you took out positive five, you would get five times negative q plus four, and the brackets would not match. Quiz and exercises now.
Your Turn
Factorise ab − 2a − 7b + 14. Check: (b − 2)(a − 7).
Watch Out
−5q + 20 is −5(q − 4), not −5(q + 4). Check the sign by expanding the pair.
📌 Key Points — this lesson in a nutshell
  • Factorising is expanding in reverse. Take out the highest common factor: the HCF of the numbers and the lowest power of each shared letter.
  • If the first term is negative, take out a negative factor — then every sign inside the bracket flips.
  • Four terms: group them in pairs, factorise each pair, and take out the matching bracket: ax + ay + bx + by = (x + y)(a + b).
  • Always check by expanding your answer. It must give back exactly what you started with.
Step 1: Introduction
Step 1 of 5
Factorise 12x + 18 1factors of 12 and 18 in common1, 2, 3, 62the highest one63divide each term by 62x and 34so12x + 18 = 6(2x + 3) Check by expanding: 6 × 2x + 6 × 3 = 12x + 18 ✓2(6x + 9) is true, but not fully factorised.Fully factorised: nothing more can come out of the bracket.
Factorise 10p3 − 15p2 1HCF of the numbers 10 and 1552lowest power of p in both termsp23HCF5p24divide each term by 5p25p2(2p − 3) Take the lowest power of each letter that is in every term.10p3 ÷ 5p2 = 2p and −15p2 ÷ 5p2 = −3Check: 5p2 × 2p − 5p2 × 3 = 10p3 − 15p2 ✓
Factorise −6a2b + 9ab2 1HCF of 6 and 9, and the letters3ab2the first term is negative: take out−3ab3−6a2b ÷ (−3ab)2a4+9ab2 ÷ (−3ab)−3b −6a2b + 9ab2 = −3ab(2a − 3b)Taking out a negative flips every sign inside.Check: −3ab × −3b = +9ab2 ✓
Factorise 3m + 3n + km + kn 1first pair: 3m + 3n3(m + n)2second pair: km + knk(m + n)3the brackets match: take (m + n) out(m + n)(3 + k)4check by expanding3m + 3n + km + kn ✓ Four terms, no factor common to all four: try pairs.The two brackets must be identical.The matching bracket is itself a common factor.
Factorise pq − 4p − 5q + 20 1first pair: pq − 4pp(q − 4)2second pair: take out −5, not 5−5(q − 4)3the brackets match(q − 4)(p − 5)4check: −5 × −4+20 ✓ If the third term is negative, take out a negative.−5q + 20 = −5(q − 4): the +20 becomes −4 inside.Quiz and exercises now.
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