Binomial expansion and binomial theorem

IB Mathematics Analysis and Approaches SL· Unit 1: Number & Algebra· 15 min read

1. Binomial Coefficients★★☆☆☆⏱ 5 min

📘 Definition

Binomial Coefficient

\binom{n}{k} = \frac{n!}{k!(n-k)!}

A value representing the number of ways to choose k items from n total items, used to find the coefficient of each term in a binomial expansion.

Example:

\binom{5}{2} = 10

Binomial coefficients can also be found using Pascal's Triangle, where the nth row (starting counting from row 0) gives all coefficients for $(a + b)^{n}$ . The first 5 rows are shown below:

\begin{aligned} \text{Row } 0: & \quad 1 \\ \text{Row } 1: & \quad 1 \quad 1 \\ \text{Row } 2: & \quad 1 \quad 2 \quad 1 \\ \text{Row } 3: & \quad 1 \quad 3 \quad 3 \quad 1 \\ \text{Row } 4: & \quad 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{aligned}

📐 Worked Example

Calculate $(3 6)$ using the factorial formula, then confirm with Pascal's triangle.

1
Substitute n=6, k=3 into the factorial formula:
2
$\binom{6}{3} = \frac{6!}{3! (6-3)!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}$
3
Cancel $3!$ from numerator and denominator, then simplify:
4
$\frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20$
5
Row 6 of Pascal's triangle is $[1, 6, 15, 20, 15, 6, 1]$ . The 4th entry (index starting at 0) is 20, which matches the calculation.

Exam tip:

For n ≤ 6, Pascal's triangle is faster than factorial calculation and less prone to arithmetic error.

2. The Binomial Theorem and Full Expansion★★☆☆☆⏱ 5 min

📘 Definition

Binomial Theorem

(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k

A general formula to expand any power of a binomial sum $(a + b)^{n}$ for positive integer n.

📐 Worked Example

Expand $(2 x - 3)^{3}$ fully using the binomial theorem.

1
Align the binomial with the formula: $a = 2 x$ , $b = - 3$ , $n = 3$ .
2
Write out each term using the binomial sum:
3
$\binom{3}{0}(2x)^3(-3)^0 + \binom{3}{1}(2x)^2(-3)^1 + \binom{3}{2}(2x)^1(-3)^2 + \binom{3}{3}(2x)^0(-3)^3$
4
Calculate each term individually:
5
1. First term: $1 \times 8 x^{3} \times 1 = 8 x^{3}$
2. Second term: $3 \times 4 x^{2} \times (- 3) = - 36 x^{2}$
3. Third term: $3 \times 2 x \times 9 = 54 x$
4. Fourth term: $1 \times 1 \times (- 27) = - 27$
6
Combine all terms for the final expansion:
7
$(2x - 3)^3 = 8x^3 - 36x^2 + 54x - 27$

3. Finding a Specific Term★★★☆☆⏱ 6 min

A very common IB exam question asks for the coefficient of a specific term (or the constant term) in a binomial expansion, without requiring full expansion. To do this, you use the general term of the expansion to solve for the k value that matches your required exponent.

For an expansion of $(a x^{m} + b)^{n}$ , the general (k+1)th term (starting from k=0) is:

T_{k+1} = \binom{n}{k} (ax^m)^{n-k} b^k

📐 Worked Example

Find the constant term in the expansion of $(x^{2} + \frac{2}{x})^{6}$ .

1
Identify values: $a = 1$ , $m = 2$ , $b = 2$ , $n = 6$ . Substitute into the general term:
2
$T_{k+1} = \binom{6}{k} (x^2)^{6-k} \left(\frac{2}{x}\right)^k$
3
Simplify the exponent of $x$ :
4
$\text{Exponent of } x = 2(6-k) - k = 12 - 3k$
5
A constant term has an exponent of 0, so solve for k:
6
$12 - 3k = 0 \implies k = 4$
7
Substitute k=4 back to find the constant term:
8
$T_{5} = \binom{6}{4} (x^2)^{2} \left(\frac{2}{x}\right)^4 = 15 \times x^4 \times \frac{16}{x^4} = 240$
9
The constant term is 240.

Exam tip:

Remember that k starts at 0, so the mth term uses k = m-1, not k=m.

4. Common Pitfalls

Wrong move:

Forgetting negative signs when expanding $(a - b)^{n}$

Why:

Treating the second term as positive b instead of -b, leading to wrong signs on odd-powered terms

Correct move:

Rewrite $(a - b)^{n}$ as $(a + (- b))^{n}$ before substitution, to explicitly track negative signs

Wrong move:

Using k equal to the term number (e.g. k=5 for the 5th term)

Why:

The first term of the expansion is k=0, so the k value is offset by 1

Correct move:

For the mth term, use k = m-1 when calculating the general term

Wrong move:

Swapping exponents: writing $(k n) a^{k} b^{n - k}$ instead of $(k n) a^{n - k} b^{k}$

Why:

Misaligning a and b with the exponents in the formula

Correct move:

Clearly label which term in your binomial is a and which is b before applying the formula

Wrong move:

Forgetting to raise the constant coefficient of a term to the correct power

Why:

Only raising the variable part to the exponent, ignoring the constant coefficient

Correct move:

Remember $(c x^{m})^{p} = c^{p} x^{m p}$ , so always calculate the coefficient's power when expanding

5. Quick Reference Cheatsheet

Concept	Formula / Rule
Binomial Coefficient	$(k n) = \frac{n !}{k ! ( n - k )!}$
Binomial Theorem	$(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k}$
General Term	$T_{k + 1} = (k n) (a x^{m})^{n - k} b^{k}$
Constant Term Condition	Set exponent of $x = 0$ , solve for $k$
Pascal's Triangle Entry	Entry = sum of two entries directly above

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 1
Find coefficient of $x^{3}$ term
2024 · Paper 1
Expand $(2 x - 3)^{4}$ fully
2023 · Paper 2
Find the constant term in expansion

Going deeper

internalCombinations and permutationsPrerequisite topic for factorial notation

What's Next

Binomial expansion is a foundational algebraic skill for IB AA SL that reappears across multiple topics, most notably in probability with the binomial distribution. The combinatorial reasoning behind binomial coefficients also underpins many counting problems for probability questions, which regularly appear on both Paper 1 and Paper 2. Even beyond exam questions, understanding the structure of binomial expansions makes expanding higher-order polynomials much faster than repeated distribution, and helps with finding derivatives or integrals of polynomial expressions. Mastery of this subtopic will also prepare you for further work on sequences and series in later units.