Math // Algebra

Factorising Expressions

Factorise algebraic expressions by taking out common factors — identify the highest common factor of all terms and write the expression as a product, e.g., 6x + 9 = 3(2x + 3)

Ages
12 to 13
Type
Method
Builds on
1 topic
Leads to
0 topics
Standards
2 codes

Quick check

Can the learner look at an expression like "6x + 9" and rewrite it with the common factor taken out — as "3(2x + 3)"?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • Identify the highest common factor of all terms in an expression
  • Write the factorised form as a product of the HCF and a bracket
  • Check factorisation by expanding the brackets back out

Topic // Map

Where this fits

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  • HelpfulHelps, but not essential

Before this

1 topic
  • Algebra // Ages 11 to 13

    Factorising is the reverse of expanding — students need expanding fluency first

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Factorising Expressions

Ages 12 to 13 // Method

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Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Factorising by common factors is the inverse of expanding: rewrite a sum as a product by pulling out the highest common factor (HCF) of every term, leaving a bracket with no remaining common factor, e.g. 6x+9 = 3(2x+3). Teach it alongside expanding, use a reverse area/array model (arrange the terms into a rectangle whose common side is the factor; the HCF gives the single rectangle with the LARGEST common side — the most compact block, not the long thin strip a smaller factor leaves), and always check by expanding the answer back. Signature errors are taking a factor that is not the highest (incomplete factorisation) and dropping or mis-dividing a term. Extend 'common factor' to a shared letter, e.g. x²+3x = x(x+3).

Common misconceptions

  • Takes a common factor that is not the HIGHEST: 12x+8 → 2(6x+4) instead of 4(3x+2) — 'factorised' but not fully.
  • Drops or fails to divide a term: 6x+9 → 3(2x) or 3(2x+9) (the 9 not divided).
  • Only spots numeric factors and misses a common variable: x²+3x left unfactorised or written 3(x²)... .
  • Does not check by expanding, so an incorrect factorisation goes unnoticed.
  • Confuses factorising (→ product) with simplifying/collecting (→ fewer terms).

Useful representations

  • Reverse area/array model (terms arranged into a rectangle; common side = factor)
  • HCF via prime factors or factor lists of the coefficients
  • Expand-back check (product → sum)
  • Side-by-side expand/factorise pair showing the inverse relationship

Teaching practices

  • Teach it as the reverse of expanding — do an expansion, then run it backwards — and always CHECK by expanding the factorised form.
  • Find the HCF of the coefficients (and any common variable) first; stress 'fully' means no common factor is left inside the bracket.
  • Use the reverse area/array model: the terms tile a rectangle whose common side is the factor; the HCF gives the neatest single rectangle.
  • Divide EVERY term by the factor — 6x÷3 and 9÷3 — and write each result inside the bracket; a dropped term is the classic slip.
  • Extend to a common variable factor, e.g. x²+3x = x(x+3), so 'common factor' includes letters, not just numbers.

Activities to try

  1. 1

    Build the rectangle with the largest common side: the reverse area model

    Why it works: The highest common factor is 'the biggest common side that leaves nothing left to share', so fully factorised means no common factor remains inside the bracket; factorising and expanding are inverse moves.

    1. Take the tiles for 12x + 8 (twelve x-tiles and eight unit tiles) and try to arrange them into one solid rectangle.
    2. First try a common side of 2: notice the result is a long 2-by-(6x+4) strip. Check whether each row could be split again into equal parts — it can, because 6x and 4 both still divide by 2, so the strip can be regrouped into a taller rectangle. That means 2 is a common factor but not the highest, so 2(6x+4) is not fully factorised.
    3. Now try a common side of 4: the tiles form a compact 4-by-(3x+2) block with nothing left to share, giving 12x + 8 = 4(3x + 2).
    4. Divide EVERY original term by the factor you pulled out (12x÷4 = 3x and 8÷4 = 2) and write each result inside the bracket.
    5. Use an 'expand back' action to re-spread the rectangle into 12x + 8 and confirm the product equals what you started with.
  2. 2

    Factorise then expand-back: the routine check

    Why it works: A factorisation is only complete when the bracket shares no further common factor, and expanding back is the way to catch dropped or mis-divided terms.

    1. Given 6x + 9, find the HCF of 6 and 9 using prime factors or factor lists, then write 6x+9 = 3(2x+3), dividing each term by 3.
    2. Immediately expand your product back out and confirm it returns the original expression.
    3. Decide whether a given bracket is fully factorised: is 2(6x+4) fully factorised? Explain how you can tell.
    4. Explain in your own words what factorising is 'undoing'.
  3. 3

    Annotated worked example: where each number comes from

    Why it works: Every term in the sum maps to factor × a term in the bracket, which is why no term may be dropped or left undivided.

    1. Study the worked line 6x + 9 = 3(2x + 3) and reveal, one piece at a time, where the 3, the 2x and the 3 inside the bracket each come from (3 is the HCF; 2x = 6x÷3; 3 = 9÷3).
    2. Trace each original term to its matching entry inside the bracket, confirming no term was dropped.

Topic // Standards

Curriculum standards

Codes for the curriculum statements this topic maps to. Each publisher owns its standard text.
Common Core State Standards for Mathematics
7.EE.1
The national curriculum in England: Key stages 1 and 2 framework document
KS3.Maths.Alg.4c

Math // Algebra