Math // Algebra

Expanding Single Brackets

Expand (multiply out) a single term over a bracket using the distributive property, e.g., 3(2x + 5) = 6x + 15; expand expressions involving negative multipliers

Ages
11 to 13
Type
Method
Builds on
1 topic
Leads to
3 topics
Standards
3 codes

Quick check

Can the learner expand a bracket like 3(2x + 5) by multiplying everything inside, to get 6x + 15?

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.
  • Multiply a single positive term over a bracket to expand the expression
  • Multiply a negative term over a bracket, correctly handling signs
  • Combine expanding brackets with collecting like terms to simplify fully

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

1 topic
  • Algebra // Ages 11 to 12

    Expanding brackets produces terms that then need collecting

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Expanding Single Brackets

Ages 11 to 13 // Method

Leads to

3 topics
  • Algebra // Ages 12 to 13

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

  • Algebra // Ages 13 to 14

    Expanding double brackets extends single-bracket expansion using FOIL/grid method

  • Algebra // Ages 11 to 14

    Many equations require expanding brackets before solving

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

Teach expanding a single bracket as the distributive law: the multiplier outside acts on EVERY term inside, and its sign travels with it. An area/grid model makes this visible — a rectangle whose width is the multiplier and whose length is the sum inside splits into one region per inside term, so 3(2x+5) is seen as 3×2x plus 3×5. Target the two persistent errors directly: distributing over only the first term (3(2x+5)→6x+5) and mishandling the sign of a negative multiplier (−2(x−4)→−2x−8, instead of the correct −2x+8). A substitution check and factoring back give learners independent ways to verify.

Common misconceptions

  • Multiplies only the first term inside: 3(2x+5) → 6x+5 (documented as 2(n+3)=2n+3 being judged correct).
  • Sign error with a negative multiplier: −2(x−4) → −2x−8 (fails to flip the second sign).
  • Confuses × with +: 3(2x+5) → 3+2x+5, or adds the outside number to the coefficient (3×2x → 5x).
  • Leaves the expression unexpanded or unsimplified after multiplying.

Useful representations

  • Area/grid rectangle split into one region per inside term
  • Arrows fanning from the multiplier to each term
  • Number substitution check (evaluate both forms at x=1)
  • Signed-multiplier annotation (−2 × −4 = +8)

Teaching practices

  • Use the area/grid model: the multiplier is one side, each inside term a separate region, so every term must be multiplied.
  • Draw an arrow from the multiplier to EACH inside term to force full distribution.
  • Handle the sign explicitly — write −2(x−4) as (−2)×x + (−2)×(−4); a negative times a negative gives +8.
  • Check by substituting a value: 3(2x+5) and 6x+15 must agree at, say, x=1 (both 21).
  • Then combine with collecting like terms, e.g. 3 + 2(x+1) → 3 + 2x + 2 → 2x + 5, stressing that only the bracket is expanded first (order of operations).

Activities to try

  1. 1

    Split the rectangle: area model of a bracket

    Why it works: Seeing the bracket as TWO regions answers 'why does the +5 not just come along unchanged?' — the multiplier covers every region, so every inside term must be multiplied, and the sign of a negative multiplier is applied to each term.

    1. Draw (or manipulate) a rectangle labelled width 3 and length (2x + 5).
    2. Split the length at the '+' into two parts, producing two regions: a 3-by-2x region and a 3-by-5 region.
    3. Label each region with its own area: 3×2x = 6x and 3×5 = 15.
    4. Gather the two areas to write the total: 3(2x+5) = 6x + 15.
    5. Now take a new example with a negative multiplier, −2(x−4), and split it into one part for each term inside the bracket. A length of −2 is not a real side length, so treat the grid as a signed diagram, not a literal area: write it as (−2)×x + (−2)×(−4), reasoning that a negative times a negative gives +8, so the result is −2x + 8.
  2. 2

    Expand, then check by substitution

    Why it works: Substituting a value gives a learner an independent way to confirm or refute their expansion, so correctness is not dependent on the teacher's say-so; an expansion and its original must agree for every value.

    1. Expand 3(2x+5) and write the result (6x + 15).
    2. Choose a value, say x=1, and evaluate both the original and the expansion: 3(2×1+5)=3×7=21 and 6×1+15=21 — they agree.
    3. Expand −2(x−4) and check at x=1: −2(1−4)=−2×−3=6 and −2×1+8=6 — they agree.
    4. Try to verify a second way by factoring the answer back to the bracket form.
    5. Decide whether each expansion is correct, correcting any that disagree.
  3. 3

    Expand within a larger expression (order of operations)

    Why it works: Only the multiplier attached to the bracket is distributed; other terms are combined afterwards by collecting like terms, respecting the order of operations.

    1. Take 3 + 2(x+1) and expand only the bracket first: 3 + 2x + 2.
    2. Collect like terms to simplify fully: 2x + 5.
    3. Explain why the +3 is not multiplied by 2 — only the term attached to the bracket is expanded first.
    4. Check the simplified form by substituting a value into both the original and the final expression.

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
6.EE.37.EE.1
The national curriculum in England: Key stages 1 and 2 framework document
KS3.Maths.Alg.4b

Math // Algebra