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Introduction to Binomial Expansion Revision Notes | GCSE AQA Higher Further Maths

To expand (a + b)ⁿ, take coefficients from row n of Pascal's triangle, with decreasing powers of a and increasing powers of b. Notes and 13 questions answered.

Updated 14 Mar 2026 · Revise Right Team

Introduction to Binomial Expansion is part of Expand (a + b)^n for positive integer n in GCSE AQA Higher Further Maths.

4 min

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24

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GCSE AQA Higher

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What this lesson covers

To expand (a + b)ⁿ, take coefficients from row n of Pascal's triangle, with decreasing powers of a and increasing powers of b. Notes and 13 questions answered.

  • Use Pascal's triangle to find the coefficients in a binomial expansion
  • Expand $(a + b)^n$ for positive integer values of $n$
  • Apply binomial expansion to expressions with numerical or algebraic terms
  • Identify and correct common errors in binomial expansions

Key ideas to keep in view

Use this lesson to stay inside Expand (a + b)^n for positive integer n while connecting the detail back to the wider Algebra topic.

  • Algebra
  • Expand (a + b)^n for positive integer n
  • GCSE AQA Higher Further Maths

Revision notes

Read the core explanations from Introduction to Binomial Expansion before testing yourself from memory.

1. Why Binomial Expansion Matters

Expanding $(x + 3)^2$ by hand is manageable. But what about $(x + 3)^5$? Multiplying out five brackets one at a time is slow, error-prone, and costs marks under exam pressure. Binomial expansion gives you a direct route to the answer.

The key insight is this: the coefficients in any expansion of $(a + b)^n$ follow a predictable pattern — one you can read straight off Pascal's triangle. Once you know the pattern, expanding any power becomes a structured, reliable process.

2. Pascal's Triangle and Binomial Coefficients

Pascal's triangle is built row by row. Each row starts and ends with 1, and every other entry is the sum of the two numbers directly above it. The row you use depends on the power $n$ you are expanding.

To extend the triangle, take any two adjacent entries and add them. For example, in row 4 the entries $4$ and $6$ are adjacent, so the entry below them in row 5 is $4 + 6 = 10$. The 1s at each end never change.

3. The Structure of a Binomial Expansion

When you expand $(a + b)^n$, the result is a sum of terms. Each term has three components: a coefficient from Pascal's triangle, a decreasing power of $a$, and an increasing power of $b$.

Notice the symmetry: the coefficients read the same forwards and backwards. That is a useful self-check — if your coefficients are not symmetric, revisit your row of Pascal's triangle.

Worked example: Expanding $(1 + x)^4$

Expand $(1 + x)^4$, giving each term in simplified form.

  1. $n = 4$, so use row 4: $1, \; 4, \; 6, \; 4, \; 1$ — There are 5 coefficients because row 4 has $4 + 1 = 5$ entries. [M1 — correct row identified]
  2. $a = 1$, $b = x$. Powers of $a$: $1^4, 1^3, 1^2, 1^1, 1^0$. Powers of $b$: $x^0, x^1, x^2, x^3, x^4$. — Powers of $a$ decrease from 4 to 0; powers of $b$ increase from 0 to 4.
  3. $1 \cdot 1^4 x^0 + 4 \cdot 1^3 x^1 + 6 \cdot 1^2 x^2 + 4 \cdot 1^1 x^3 + 1 \cdot 1^0 x^4$ — Each coefficient from Pascal's triangle multiplies the corresponding power pair. [M1 — correct structure]
  4. $1 + 4x + 6x^2 + 4x^3 + x^4$ — $1^k = 1$ for any $k$, so those factors disappear. [A1 — fully simplified expansion]
  5. Term 1: $0+4=4$ ✓, Term 2: $1+3=4$ ✓, Term 3: $2+2=4$ ✓, Term 4: $3+1=4$ ✓, Term 5: $4+0=4$ ✓ — Every pair of powers sums to $n = 4$. Coefficients $1,4,6,4,1$ are symmetric. Both checks pass.

Answer: $(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$

4. Expanding with Numerical and Algebraic Terms

The method extends directly when $a$ or $b$ is a number rather than 1. The key extra step is raising each numerical term to its required power before multiplying by the Pascal coefficient.

Worked example: Expanding $(2 + x)^3$

Expand $(2 + x)^3$, giving each term in simplified form.

  1. $n = 3$, so use row 3: $1, \; 3, \; 3, \; 1$ — Four coefficients for four terms. [M1 — correct row]
  2. $1 \cdot 2^3 x^0 + 3 \cdot 2^2 x^1 + 3 \cdot 2^1 x^2 + 1 \cdot 2^0 x^3$ — Here $a = 2$ and $b = x$. The power of 2 decreases; the power of $x$ increases. [M1 — correct structure]
  3. $2^3 = 2 \times 2 \times 2 = 8$, $\quad 2^2 = 2 \times 2 = 4$, $\quad 2^1 = 2$, $\quad 2^0 = 1$ — Expand each power of 2 longhand to avoid arithmetic errors.
  4. $1 \times 8 + 3 \times 4 x + 3 \times 2 x^2 + 1 \times 1 x^3$ — Pascal coefficient multiplied by the evaluated power of 2.
  5. $8 + 12x + 6x^2 + x^3$ — [A1 — all four terms correct and simplified]

Answer: $(2 + x)^3 = 8 + 12x + 6x^2 + x^3$

The same approach works when both terms are algebraic. For $(x + 2y)^4$, treat $a = x$ and $b = 2y$, then raise $2y$ to each required power in turn.

Worked example: Expanding $(x + 2y)^4$

Expand $(x + 2y)^4$, giving each term in simplified form.

  1. $n = 4$, row 4: $1, \; 4, \; 6, \; 4, \; 1$ — Five terms in the expansion.
  2. $1 \cdot x^4(2y)^0 + 4 \cdot x^3(2y)^1 + 6 \cdot x^2(2y)^2 + 4 \cdot x^1(2y)^3 + 1 \cdot x^0(2y)^4$ — $a = x$, $b = 2y$. Treat $(2y)$ as a single unit when raising to each power. [M1]
  3. $(2y)^0 = 1$, $\quad (2y)^1 = 2y$, $\quad (2y)^2 = 4y^2$, $\quad (2y)^3 = 8y^3$, $\quad (2y)^4 = 16y^4$ — $(2y)^2 = 2^2 \times y^2 = 4y^2$. $(2y)^3 = 2^3 \times y^3 = 8y^3$. $(2y)^4 = 2^4 \times y^4 = 16y^4$. Raise both the number and the variable.
  4. $1 \cdot x^4 \cdot 1 + 4 \cdot x^3 \cdot 2y + 6 \cdot x^2 \cdot 4y^2 + 4 \cdot x \cdot 8y^3 + 1 \cdot 1 \cdot 16y^4$ — Combine the three factors in each term.
  5. $x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4$ — $4 \times 2 = 8$, $6 \times 4 = 24$, $4 \times 8 = 32$. [A1 — all five terms correct]

Answer: $(x + 2y)^4 = x^4 + 8x^3y + 24x^2y^2 + 32xy^3 + 16y^4$

5. Expanding with a Negative Second Term

When the second term is negative — for example $(x - 3)^3$ — you rewrite it as $(x + (-3))^3$ and apply the same method. The signs of the terms in the expansion will alternate: even powers of a negative number are positive and odd powers are negative, so the signs cycle $+, -, +, -$ throughout.

Worked example: Expanding $(x - 3)^3$

Expand $(x - 3)^3$, giving each term in simplified form.

  1. $(x + (-3))^3$, so $a = x$, $b = -3$ — Treating $b$ as $-3$ ensures signs are handled correctly throughout.
  2. $1, \; 3, \; 3, \; 1$ — $n = 3$, so use row 3.
  3. $1 \cdot x^3(-3)^0 + 3 \cdot x^2(-3)^1 + 3 \cdot x^1(-3)^2 + 1 \cdot x^0(-3)^3$ — Raise $-3$ (not just 3) to each power. [M1]
  4. $(-3)^0 = 1$, $\quad (-3)^1 = -3$, $\quad (-3)^2 = (-3) \times (-3) = 9$, $\quad (-3)^3 = (-3) \times (-3) \times (-3) = -27$ — Even powers give positive results; odd powers give negative results.
  5. $x^3 + 3 \cdot x^2 \cdot (-3) + 3 \cdot x \cdot 9 + 1 \cdot (-27)$ $= x^3 - 9x^2 + 27x - 27$ — $3 \times (-3) = -9$, $3 \times 9 = 27$. Signs alternate: $+, -, +, -$. [A1 — all terms correct]

Answer: $(x - 3)^3 = x^3 - 9x^2 + 27x - 27$

6. Practice Problems

The method is consistent across all cases. Work through these problems using the same structured approach: identify the row, set up the terms, evaluate powers, then simplify.

7. Exam Technique and Common Mistakes

Binomial expansion questions reward method marks heavily. Even if you make an arithmetic slip, you can still earn marks for using the correct row of Pascal's triangle and setting up the term structure correctly.

The most common source of lost marks is sign errors when the second term is negative. Writing $b = -3$ and then computing $(-3)^2 = 9$ (positive) is correct — but many students write $-3^2 = -9$ by mistake. Always raise the entire term, including its sign, to the required power.

Binomial expansion at GCSE centres on two skills: building Pascal's triangle accurately and applying the coefficient-power structure to any $(a + b)^n$ with positive integer $n$. Master those two steps and the method works for every question the exam can set.

Frequently asked questions

How do you expand (a + b)ⁿ?

To expand (a + b)ⁿ, take the coefficients from row n of Pascal's triangle, then write each term with a decreasing power of a and an increasing power of b. Each pair of powers sums to n, and row n contains exactly n + 1 entries, so there are n + 1 terms.

What do you need to know about binomial expansion for GCSE Further Maths?

Binomial expansion at GCSE centres on two skills: building Pascal's triangle accurately and applying the coefficient-power structure to any (a + b)ⁿ with positive integer n. It includes expansions with numerical terms, algebraic terms such as (x + 2y)⁴, and negative terms such as (x − 3)³.

How does Pascal's triangle work?

Pascal's triangle is built row by row: each row starts and ends with 1, and every other entry is the sum of the two adjacent entries in the row before. For example, the adjacent entries 4 and 6 in row 4 give 10 in row 5. Row n gives the coefficients for (a + b)ⁿ.

How do you expand (1 + x)⁴?

Use row 4 of Pascal's triangle, 1, 4, 6, 4, 1, with powers of 1 decreasing and powers of x increasing. Since every power of 1 equals 1, (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴.

How do you expand a binomial with a number term?

Raise each numerical term to its required power before multiplying by the Pascal coefficient. For (2 + x)³, row 3 is 1, 3, 3, 1 and the powers of 2 are 8, 4, 2, 1, so (2 + x)³ = 8 + 12x + 6x² + x³.

How do you expand a binomial with a negative term?

Rewrite the bracket with a negative b and raise the whole term, including its sign, to each power. For (x − 3)³, b = −3, so the powers are 1, −3, 9 and −27, giving x³ − 9x² + 27x − 27, with signs alternating +, −, +, −.

How do you expand binomials at GCSE?

Binomials such as (a + b)ⁿ are expanded in four steps: identify row n of Pascal's triangle, set up each term as coefficient × decreasing power of a × increasing power of b, evaluate the powers, then simplify. For example, (1 + x)⁴ = 1 + 4x + 6x² + 4x³ + x⁴.

Is binomial expansion in GCSE?

Binomial expansion appears at GCSE in the form of expanding (a + b)ⁿ for positive integer n, using Pascal's triangle to find the coefficients. Expanding (x + 3)⁵ bracket by bracket is slow and error-prone, so this method gives a direct route to the answer.

How many terms are in the expansion of (a + b)ⁿ?

The expansion of (a + b)ⁿ has n + 1 terms, because row n of Pascal's triangle contains exactly n + 1 entries. For example, (1 + x)⁴ has 5 terms, with coefficients 1, 4, 6, 4, 1.

How do you expand (2 + x)³?

(2 + x)³ = 8 + 12x + 6x² + x³. Using row 3 of Pascal's triangle (1, 3, 3, 1) with powers of 2 decreasing, the terms are 1 × 8, 3 × 4x, 3 × 2x² and 1 × x³.

How do you expand (x + 2y)⁴?

(x + 2y)⁴ = x⁴ + 8x³y + 24x²y² + 32xy³ + 16y⁴. Row 4 of Pascal's triangle gives 1, 4, 6, 4, 1, and 2y is treated as a single unit, so (2y)² = 4y², (2y)³ = 8y³ and (2y)⁴ = 16y⁴.

How do you check a binomial expansion?

A binomial expansion is checked in two ways: every pair of powers in each term should sum to n, and the coefficients from Pascal's triangle should be symmetric, reading the same forwards and backwards. If the coefficients are not symmetric, the row of Pascal's triangle needs revisiting.

Why is (−3)² not the same as −3²?

(−3)² = (−3) × (−3) = 9, while −3² = −9, so raising only the number and not its sign is a common mistake in binomial expansion. When the second term is negative, the entire term, including its sign, must be raised to each power.

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