# 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

Example 1

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}

Example 2

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.

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

Example 3

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.

Example 4

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 $$\ (a+b)^{4}$$.

$$\ 1+4+6+4+1=16$$

Example 5

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.

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

Compute the following summations.

1. $$\ \sum_{i=0}^{9}\left(\begin{array}{l} 9 \\ i \end{array}\right)$$
2. $$\ \sum_{i=0}^{12}\left(\begin{array}{c} 12 \\ i \end{array}\right)$$
3. $$\ \sum_{i=0}^{8}\left(\begin{array}{l} 8 \\ i \end{array}\right)$$

Collapse the following polynomials using the Binomial Theorem.

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

# Vocabulary

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