Expanding Brackets

Expanding brackets is a key algebra skill in GCSE Maths used to simplify algebraic expressions and solve equations. Expanding brackets means multiplying everything inside the bracket by the term outside. In this lesson, we will learn how to expand single and double brackets step by step using clear examples and practice questions.

What does expanding brackets mean?

Expanding brackets removes the bracket by multiplying each term inside it.

Examples:

3(x+4)3(x + 4)

2(a−5)2(a − 5)

x(x+3)x(x + 3)

We multiply the outside term by every term inside the bracket.

Example Question:

Expand:

3(x+4)3(x + 4)

Exam Tip

Always multiply every term inside the bracket — not just the first one.

Missing one term is the most common mistake in exams.

Look for:

  • Single brackets
  • Negative signs
  • Terms with variables

Method – Expanding Single Brackets

To expand a bracket:

  1. Multiply the outside term by the first term inside
  2. Multiply the outside term by the second term inside
  3. Write the results and simplify if possible

Worked Example

Expand:

3(x+4)3(x + 4)

Step 1: Multiply 33 by xx

3×x=3x3 × x = 3x

Step 2: Multiply 33 by 44

3×4=123 × 4 = 12

Step 3: Write the result

3x+123x + 12

Worked Example 2

Expand:

5(a−2)5(a − 2)

Step 1: Multiply 55 by aa

5a5a

Step 2: Multiply 55 by −2−2

5×−2=−105 × −2 = −10

Step 3: Write the result

5a−105a − 10

Worked Example 3 (With Variables)

Expand:

x(x+3)x(x + 3)

Step 1: Multiply xx by xx

x×x=x2x × x = x²

Step 2: Multiply xx by 33

3x3x

Step 3: Write the result

x2+3xx² + 3x

Worked Example 4 (Negative Outside)

Expand:

−2(x+5)−2(x + 5)

Step 1: Multiply −2−2 by xx

−2x−2x

Step 2: Multiply −2−2 by 55

−10−10

Step 3: Write the result

−2x−10−2x − 10

Expanding double brackets

Multiply every term in the first bracket by every term in the second bracket. For two terms in each bracket, this gives four products. Keep the signs with the terms, then collect like terms.

Worked Example 5 (Two positive terms)

Expand and simplify: (x + 3)(x + 2)

Step 1: Multiply x by both terms: x × x + x × 2 = x² + 2x.

Step 2: Multiply 3 by both terms: 3 × x + 3 × 2 = 3x + 6.

Step 3: Combine the four products: x² + 2x + 3x + 6.

Answer: x² + 5x + 6. Only the 2x and 3x terms combine.

Worked Example 6 (A coefficient and a negative term)

Expand and simplify: (2x − 1)(x + 4)

Step 1: Multiply 2x by both terms: 2x × x + 2x × 4 = 2x² + 8x.

Step 2: Multiply −1 by both terms: −1 × x + (−1) × 4 = −x − 4.

Step 3: Collect like terms: 2x² + 8x − x − 4.

Answer: 2x² + 7x − 4. Remember that the negative sign belongs to the −1.

Final answer:

The expanded expression contains no brackets and is fully simplified.

Check your solution (important habit)

Ensure:

• Every term inside the bracket was multiplied

• Signs are correct

• Like terms are simplified if needed

Quick check example:

2(x+3)→2x+62(x + 3) → 2x + 6 ✔️

2(x+3)→2x+32(x + 3) → 2x + 3 ❌

Combining brackets and extending the method

Worked example — Two separate brackets

Expand and simplify 3(x + 2) − 2(x − 4).

3x + 6 − 2x + 8

x + 14.

The −2 multiplies both terms in its bracket. In particular, (−2) × (−4) = +8.

Worked example — Squaring a bracket

(x + 4)² means (x + 4)(x + 4). It does not mean x² + 16.

Multiply all four pairs: x² + 4x + 4x + 16.

Collect the middle terms: x² + 8x + 16.

Higher extension — Three brackets

Expand (x + 1)(x + 2)(x − 1). Work with two brackets first.

(x + 1)(x − 1) = x² − x + x − 1 = x² − 1.

Now multiply by x + 2: (x² − 1)(x + 2).

x³ + 2x² − x − 2.

The final answer is x³ + 2x² − x − 2. There are no like terms left to collect.

Identities, arguments and proof

An identity holds for every allowed value. For example, (x + 2)² and x² + 4x + 4 are equivalent because expanding the first gives the second. The equation x + 2 = 5 holds only at x = 3. Explaining equivalent expressions is a Foundation and Higher skill; the general proofs below are Higher.

Higher worked example — Prove a statement for every odd integer

Prove that the square of any odd integer is one more than a multiple of 8. Write an arbitrary odd integer as 2n + 1, where n is an integer.

(2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.

The consecutive integers n and n + 1 include an even number, so n(n + 1) = 2k for some integer k. The square is therefore 8k + 1. This proves the claim for every odd integer, including negative ones.

Higher worked example — A counterexample disproves an “always” claim

Claim: n² + n + 1 is prime for every positive integer n. Taking n = 4 gives 16 + 4 + 1 = 21 = 3 × 7. This single counterexample disproves the claim. Several examples that work would not prove it.

Higher practice — Prove that the difference between the squares of consecutive integers is odd.

Let the integers be n and n + 1. Then (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1.

Since n is an integer, 2n + 1 is odd. This is a general argument, not a check of selected numbers.

Higher practice — Disprove the claim that (a + b)² = a² + b² for every a and b.

For a = 1 and b = 1, the left side is 4 and the right side is 2. They differ, so the claim is false.

In general, expanding gives (a + b)² = a² + 2ab + b²; the proposed rule loses the middle term.

Think Like a Mathematician

Expanding brackets is based on the distributive law.

This means multiplication is “distributed” across addition or subtraction inside the bracket.

Understanding this helps when working with more complex algebra later.

Common mistakes to avoid

Forgetting to multiply the second term

• Getting negative signs wrong

• Not simplifying the result

• Mixing up multiplication and addition

Negative numbers cause most errors — slow down and check signs carefully.

Try one yourself

Expand:

4(x−3)4(x − 3)

Multiply 4 by each term inside the bracket.

4 × x = 4x and 4 × (−3) = −12.

Answer: 4x − 12.

Now try double brackets: expand and simplify (x − 2)(x + 5).

Multiply each term in the first bracket by both terms in the second.

(x − 2)(x + 5) = x² + 5x − 2x − 10.

Collect the x terms: 5x − 2x = 3x.

Answer: x² + 3x − 10.

More practice — expanding

1. Expand −3(2x − 5).

(−3) × 2x = −6x and (−3) × (−5) = 15.

−6x + 15.

2. Expand and simplify 2(x + 5) − 3(x − 1).

2x + 10 − 3x + 3.

Collect like terms: 13 − x.

3. Expand (x − 3)².

Write (x − 3)(x − 3).

x² − 3x − 3x + 9 = x² − 6x + 9.

4. Higher: expand (x − 2)(x + 2)(x + 3).

The first pair gives x² − 4.

(x² − 4)(x + 3) = x³ + 3x² − 4x − 12.

Frequently Asked Questions

What does expanding brackets mean?

Expanding brackets means multiplying the terms inside the brackets by the term outside to remove the brackets.

What is the difference between single and double brackets?

Single brackets involve multiplying one term by several terms, while double brackets involve multiplying two expressions together.

Why does x×xx × x become x2x²?

Because multiplying a variable by itself gives the variable squared.

x×x=x2x × x = x²

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