# 7.5.1: Binomial Theorem and Expansions

- Page ID
- 14798

## Binomial Theorem

The Binomial Theorem tells you how to expand a binomial such as \(\ (2 x-3)^{5}\) without having to compute the repeated distribution. What is the expanded version of \(\ (2 x-3)^{5}\)?

## Introduction to the Binomial Theorem

The **Binomial Theorem** states:

\(\ (a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}

n \\

i

\end{array}\right) a^{i} b^{n-i}\)

Writing out a few terms of the summation symbol helps you to understand how this theorem works:

\(\ (a+b)^{n}=\left(\begin{array}{c}

n \\

0

\end{array}\right) a^{n}+\left(\begin{array}{c}

n \\

1

\end{array}\right) a^{n-1} b^{1}+\left(\begin{array}{c}

n \\

2

\end{array}\right) a^{n-2} b^{2}+\cdots+\left(\begin{array}{c}

n \\

n

\end{array}\right) b^{n}\)

Going from one term to the next in the expansion, you should notice that the exponents of \(\ a\) decrease while the exponents of \(\ b\) increase. You should also notice that the coefficients of each term are combinations. Recall that \(\ \left(\begin{array}{l}

n \\

0

\end{array}\right)\) is the number of ways to choose objects from a set of \(\ n\) objects.

Take the following binomial:

\(\ (m-n)^{6}\)

It can be expanded using the Binomial Theorem:

\(\ \begin{aligned}

(m-n)^{6}=&\left(\begin{array}{c}

6 \\

0

\end{array}\right) m^{6}+\left(\begin{array}{c}

6 \\

1

\end{array}\right) m^{5}(-n)^{1}+\left(\begin{array}{c}

6 \\

2

\end{array}\right) m^{4}(-n)^{2}+\left(\begin{array}{c}

6 \\

3

\end{array}\right) m^{3}(-n)^{3} \\

&+\left(\begin{array}{c}

6 \\

4

\end{array}\right) m^{2}(-n)^{4}+\left(\begin{array}{c}

6 \\

5

\end{array}\right) m^{1}(-n)^{5}+\left(\begin{array}{c}

6 \\

6

\end{array}\right)(-n)^{6} \\

=& 1 m^{6}-6 m^{5} n+15 m^{4} n^{2}-20 m^{3} n^{3}+15 m^{2} n^{4}-6 m^{1} n^{5}+1 n^{6}

\end{aligned}\)

Be extremely careful when working with binomials of the form \(\ (a-b)^{n}\). You need to remember to capture the negative with the second term as you write out the expansion: \(\ (a-b)^{n}=(a+(-b))^{n}\).

Another way to think about the coefficients in the Binomial Theorem is that they are the numbers from Pascal’s Triangle. Look at the expansions of \(\ (a+b)^{n}\) below and notice how the coefficients of the terms are the numbers in Pascal’s Triangle.

\(\ \begin{array}{c}

(a+b)^{0}=1 \\

(a+b)^{1}=1 a+1 b \\

(a+b)^{2}=1 a^{2}+2 a b+1 b^{2} \\

(a+b)^{3}=1 a^{3}+3 a^{2} b+3 a b^{2}+1 b^{3} \\

(a+b)^{4}=1 a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+1 b^{4}\\

\vdots

\end{array}\)

## Examples

Earlier, you were asked to expand \(\ (2 x-3)^{5}\). The expanded version of \(\ (2 x-3)^{5}\) is:

**Solution**

\(\ \begin{aligned}

(2 x-3)^{5}=&\left(\begin{array}{c}

5 \\

0

\end{array}\right)(2 x)^{5}+\left(\begin{array}{c}

5 \\

1

\end{array}\right)(2 x)^{4}(-3)^{1}+\left(\begin{array}{c}

5 \\

2

\end{array}\right)(2 x)^{3}(-3)^{2} \\

&+\left(\begin{array}{c}

5 \\

3

\end{array}\right)(2 x)^{2}(-3)^{3}+\left(\begin{array}{c}

5 \\

4

\end{array}\right)(2 x)^{1}(-3)^{4}+\left(\begin{array}{c}

5 \\

5

\end{array}\right)(-3)^{6} \\

=&(2 x)^{5}+5(2 x)^{4}(-3)^{1}+10(2 x)^{3}(-3)^{2} \\

&+10(2 x)^{2}(-3)^{3}+5(2 x)^{1}(-3)^{4}+(-3)^{5} \\

=& 32 x^{5}-240 x^{4}+720 x^{3}-1080 x^{2}+810 x-243

\end{aligned}\)

What is the coefficient of the term \(\ x^{7} y^{9}\) in the expansion of the binomial \(\ (x+y)^{16}\)?

**Solution**

The Binomial Theorem allows you to calculate just the coefficient you need.

\(\ \left(\begin{array}{c}

16 \\

9

\end{array}\right)=\frac{16 !}{9 ! 7 !}=\frac{16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=11,440\)

What is the coefficient of \(\ x^{6}\) in the expansion of \(\ (4-3 x)^{7}\)?

**Solution**

For this problem you should calculate the whole term, since the 3 and the 4 in \(\ (3-4 x)\) will impact the coefficient of \(\ x^{6}\) as well. \(\ \left(\begin{array}{l}

7 \\

6

\end{array}\right) 4^{1}(-3 x)^{6}=7 \cdot 4 \cdot 729 x^{6}=20,412 x^{6}\). The coefficient is 20,412.

Compute the following summation.

\(\ \sum_{i=0}^{4}\left(\begin{array}{l}

4 \\

i

\end{array}\right)\)

**Solution**

This is asking for \(\ \left(\begin{array}{l}

4 \\

0

\end{array}\right)+\left(\begin{array}{l}

4 \\

1

\end{array}\right)+\cdots+\left(\begin{array}{l}

4 \\

4

\end{array}\right)\), which are the sum of all the coefficients of \(\ (a+b)^{4}\).

\(\ 1+4+6+4+1=16\)

Collapse the following polynomial using the Binomial Theorem.

\(\ 32 x^{5}-80 x^{4}+80 x^{3}-40 x^{2}+10 x-1\)

**Solution**

Since the last term is -1 and the power on the first term is a 5 you can conclude that the second half of the binomial is \(\ (?-1)^{5}\). The first term is positive and \(\ (2 x)^{5}=32 x^{5}\), so the first term in the binomial must be \(\ 2 x\). The binomial is \(\ (2 x-1)^{5}\).

## Review

Expand each of the following binomials using the Binomial Theorem.

- \(\ (x-y)^{4}\)
- \(\ (x-3 y)^{5}\)
- \(\ (2 x+4 y)^{7}\)
- What is the coefficient of \(\ x^{4}\) in \(\ (x-2)^{7}\)?
- What is the coefficient of \(\ x^{3} y^{5}\) in \(\ (x+y)^{8}\)?
- What is the coefficient of \(\ x^{5}\) in \(\ (2 x-5)^{6}\)?
- What is the coefficient of \(\ y^{2}\)
- What is the coefficient of \(\ x^{2} y^{6}\) in \(\ (2 x+y)^{8}\)?
- What is the coefficient of \(\ x^{3} y^{4}\) in \(\ (5 x+2 y)^{7}\)?

Compute the following summations.

- \(\ \sum_{i=0}^{9}\left(\begin{array}{l}

9 \\

i

\end{array}\right)\) - \(\ \sum_{i=0}^{12}\left(\begin{array}{c}

12 \\

i

\end{array}\right)\) - \(\ \sum_{i=0}^{8}\left(\begin{array}{l}

8 \\

i

\end{array}\right)\)

Collapse the following polynomials using the Binomial Theorem.

- \(\ 243 x^{5}-405 x^{4}+270 x^{3}-90 x^{2}=15 x-1\)
- \(\ x^{7}-7 x^{6} y+21 x^{5} y^{2}-35 x^{4} y^{3}+35 x^{3} y^{4}-21 x^{2} y^{5}+7 x y^{6}-y^{7}\)
- \(\ 128 x^{7}-448 x^{6} y+672 x^{5} y^{2}-560 x^{4} y^{3}+280 x^{3} y^{4}-84 x^{2} y^{5}+14 x y^{6}-y^{7}\)

## Vocabulary

Term | Definition |
---|---|

combination |
Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set. |

## Image Attributions

- [Figure 1]

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