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The Binomial Theorem (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 9 April 2026

Binomial theorem

What is a binomial?

A binomial is an expression consisting of two terms added together
- E.g. $(a + b)$ , $(x + 3)$
Expanding a binomial raised to a power means writing it as a polynomial in standard form
- E.g. $(x + 3)^{2} = x^{2} + 6 x + 9$

What is Pascal's Triangle?

Pascal's Triangle is a triangular arrangement of numbers where each entry is the sum of the two entries directly above it
The rows of Pascal's Triangle are numbered starting from row $0$ :
- Row 0: $1$
- Row 1: $1 1$
- Row 2: $1 2 1$
- Row 3: $1 3 3 1$
- Row 4: $1 4 6 4 1$
- Row 5: $1 5 10 10 5 1$
The table can be extended to new rows by
- adding the two numbers above each position
- and putting 1s on the outside
The entries in row $n$ are the coefficients needed to expand $(a + b)^{n}$

A diagram of Pascal's Triangle showing six rows of numbers; each row is indented so the triangle is symmetrical, ranging from 1 at the top to 10 in the middle.

How do I use Pascal's Triangle to expand (a+b)ⁿ?

To expand $(a + b)^{n}$
- Look up row $n$ of Pascal's Triangle for the coefficients
- Write out terms where for each term
  - the power of $a$ decreases from $n$ to $0$
  - and the power of $b$ increases from $0$ to $n$
    - There will be $n + 1$ terms in total
- Multiply each term by the corresponding coefficient from Pascal's Triangle
  - and then add all the terms together
- Simplify
E.g. to expand $(a + b)^{3}$ :
- Row 3 of Pascal's Triangle gives coefficients: $1, 3, 3, 1$
- The terms are $a^{3} b^{0}, a^{2} b^{1}, a^{1} b^{2}, a^{0} b^{3}$
- Multiplying by the coefficients and adding together gives
  - $(a + b)^{3} = 1 \cdot a^{3} b^{0} + 3 \cdot a^{2} b^{1} + 3 \cdot a^{1} b^{2} + 1 \cdot a^{0} b^{3}$
- Simplify, including using $a^{0} = b^{0} = 1$ , $a^{1} = a$ and $b^{1} = b$
  - $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$
Following the same procedure for $(a + b)^{4}$ gives
- $(a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}$

How do I expand polynomial functions of the form (x+c)ⁿ?

The binomial theorem applies directly to expressions of the form $(x + c)^{n}$
- where $c$ is a constant
This is just the general $(a + b)^{n}$ expansion with $a = x$ and $b = c$
- After expanding, simplify by evaluating the powers of $c$ and collecting terms
E.g. to expand $(x + 2)^{3}$ :
- Row 3 of Pascal's Triangle gives coefficients: $1, 3, 3, 1$
- The terms are $x^{3} \cdot 2^{0}, x^{2} \cdot 2^{1}, x^{1} \cdot 2^{2}, x^{0} \cdot 2^{3}$
- Multiplying by the coefficients and adding together gives
  - $(x + 2)^{3} = 1 \cdot x^{3} \cdot 2^{0} + 3 \cdot x^{2} \cdot 2^{1} + 3 \cdot x^{1} \cdot 2^{2} + 1 \cdot x^{0} \cdot 2^{3}$
- Simplify, including using $x^{0} = 2^{0} = 1$ and $x^{1} = x$
  - $\begin{array}{rcl} (x + 2)^{3} & = & 1 \cdot x^{3} \cdot 1 + 3 \cdot x^{2} \cdot 2 + 3 \cdot x \cdot 4 + 1 \cdot 1 \cdot 8 \\ = & x^{3} + 6 x^{2} + 12 x + 8 \end{array}$
Or to expand $(x - 4)^{4}$ :
- Here $b = - 4$
  - (note the negative sign, i.e. treat the subtraction as addition of a negative)
- Row 4 of Pascal's Triangle gives coefficients: $1, 4, 6, 4, 1$
- The terms are $x^{4} \cdot {(- 4)}^{0}, x^{3} \cdot {(- 4)}^{1}, x^{2} \cdot {(- 4)}^{2}, x^{1} \cdot {(- 4)}^{3}, x^{0} \cdot {(- 4)}^{4}$
- Multiplying by the coefficients and adding together gives
  - ${(x - 4)}^{4} = 1 \cdot x^{4} \cdot {(- 4)}^{0} + 4 \cdot x^{3} \cdot {(- 4)}^{1} + 6 \cdot x^{2} \cdot {(- 4)}^{2} + 4 \cdot x^{1} \cdot {(- 4)}^{3} + 1 \cdot x^{0} \cdot {(- 4)}^{4}$
  - Simplify, including using $x^{0} = {(- 4)}^{0} = 1$ and $x^{1} = x$
    - $\begin{array}{rcl} {(x - 4)}^{4} & = & 1 \cdot x^{4} \cdot 1 + 4 \cdot x^{3} \cdot (- 4) + 6 \cdot x^{2} \cdot 16 + 4 \cdot x \cdot (- 64) + 1 \cdot 1 \cdot 256 \\ = & x^{4} - 16 x^{3} + 96 x^{2} - 256 x + 256 \end{array}$

Examiner Tips and Tricks

When expanding $(x + c)^{n}$ where $c$ is negative, be very careful with signs. It can help to remember that

a negative number raised to an odd power gives a negative answer
a negative number raised to an even power gives a positive answer

Worked Example

Use Pascal's Triangle to expand $(x - 3)^{4}$ . Write your answer in the form $a x^{4} + b x^{3} + c x^{2} + d x + f$ , where $a, b, c, d$ and $f$ are integers to be found.

Answer:

Use the standard Pascal's triangle method for expanding ${(a + b)}^{n}$

with $a = x$ and $b = - 3$

Row 4 of Pascal's Triangle gives coefficients: $1, 4, 6, 4, 1$

The terms are $x^{4} \cdot {(- 3)}^{0}, x^{3} \cdot {(- 3)}^{1}, x^{2} \cdot {(- 3)}^{2}, x^{1} \cdot {(- 3)}^{3}, x^{0} \cdot {(- 3)}^{4}$
So multiplying by the coefficients and adding together gives

${(x - 3)}^{4} = 1 \cdot x^{4} \cdot {(- 3)}^{0} + 4 \cdot x^{3} \cdot {(- 3)}^{1} + 6 \cdot x^{2} \cdot {(- 3)}^{2} + 4 \cdot x^{1} \cdot {(- 3)}^{3} + 1 \cdot x^{0} \cdot {(- 3)}^{4}$

Simplify, including using $x^{0} = {(- 3)}^{0} = 1$ and $x^{1} = x$

Be careful calculating the powers of $- 3$
- $(- 3)^{0} = 1$ , $(- 3)^{1} = - 3$ , $(- 3)^{2} = 9$ , $(- 3)^{3} = - 27$ , and $(- 3)^{4} = 81$
- So

$\begin{array}{rcl} {(x - 3)}^{4} & = & 1 \cdot x^{4} \cdot 1 + 4 \cdot x^{3} \cdot (- 3) + 6 \cdot x^{2} \cdot 9 + 4 \cdot x^{1} \cdot (- 27) + 1 \cdot x^{0} \cdot 81 \\ = & x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81 \end{array}$

That is in the form required, with $a = 1$ , $b = - 12$ , $c = 54$ , $d = - 108$ and $f = 81$

$\begin{array}{rcl} (x - 3) \end{array} \begin{array}{rcl} \end{array}$

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The Binomial Theorem (College Board AP® Precalculus): Study Guide

Binomial theorem

What is a binomial?

What is Pascal's Triangle?

How do I use Pascal's Triangle to expand (a+b)ⁿ?

How do I expand polynomial functions of the form (x+c)ⁿ?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis

The Binomial Theorem (College Board AP® Precalculus): Study Guide

Binomial theorem

What is a binomial?

What is Pascal's Triangle?

How do I use Pascal's Triangle to expand (a+b)n?

How do I expand polynomial functions of the form (x+c)n?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis

How do I use Pascal's Triangle to expand (a+b)ⁿ?

How do I expand polynomial functions of the form (x+c)ⁿ?