9.4: Binomial Theorem

Learning Objectives

• Evaluate expressions involving factorials.
• Calculate binomial coefficients.
• Expand powers of binomials using the binomial theorem.

Factorials and the Binomial Coefficient

We begin by defining the factorial²⁵ of a natural number $n$, denoted $n!$, as the product of all natural numbers less than or equal to $n$.

$n !=n(n-1)(n-2) \cdots 3 \cdot 2 \cdot 1$

For example,

$\begin{array}{l}{7 !=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5,040 \quad\color{Cerulean}{Seven\:factorial}} \\[4pt] {5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=120\quad\quad\quad\quad\:\color{Cerulean}{Five\:factorial}} \\[4pt] {3 !=3 \cdot 2 \cdot 1=6\quad\quad\quad\quad\quad\quad\:\:\:\quad\color{Cerulean}{Three\:factorial}} \\[4pt] {1 !=1=1\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\color{Cerulean}{One\:factorial}}\end{array}$

We define zero factorial²⁶ to be equal to $1$,

$0 !=1 \quad\color{Cerulean}{Zero\:factorial}$

The factorial of a negative number is not defined.

On most modern calculators you will find a factorial function. Some calculators do not provide a button dedicated to it. However, it usually can be found in the menu system if one is provided.

The factorial can also be expressed using the following recurrence relation,

$n !=n(n-1) !$

For example, the factorial of $8$ can be expressed as the product of $8$ and $7!$:

$\begin{aligned} 8 ! &=8 \cdot \color{Cerulean}{7 !} \\[4pt] &=8 \cdot \color{Cerulean}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\[4pt] &=40,320 \end{aligned}$

When working with ratios involving factorials, it is often the case that many of the factors cancel.

Example $\PageIndex{1}$

Evaluate: $\frac{12 !}{6 !}$.

Solution

$\begin{aligned} \frac{12 !}{6 !}&=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot \color{Cerulean}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{\color{Cerulean}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}\\[4pt] &=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7 \cdot \cancel{6!}}{\cancel{6!}} \\[4pt] &=12\cdot11\cdot10\cdot9\cdot8\cdot7\\[4pt] &=665,280\end{aligned}$

Answer

$665,280$

The binomial coefficient²⁷, denoted $_{n} C_{k}=\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)$, is read “$n$ choose $k$” and is given by the following formula:

$_{n} C_{k}=\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$

This formula is very important in a branch of mathematics called combinatorics. It gives the number of ways $k$ elements can be chosen from a set of $n$ elements where order does not matter. In this section, we are concerned with the ability to calculate this quantity.

Example $\PageIndex{2}$

Calculate $\left( \begin{array}{l}{7} \\[4pt] {3}\end{array}\right)$.

Solution

Use the formula for the binomial coefficient,

$\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$

where $n = 7$ and $k = 3$. After substituting, look for factors to cancel.

$\begin{aligned} \left( \begin{array}{l}{7} \\[4pt] {3}\end{array}\right) &=\frac{7 !}{3 !(7-3) !} \\[4pt] &=\frac{7 !}{3 ! 4 !} \\[4pt] &=\frac{7\cdot6\cdot5\cdot\cancel{\color{Cerulean}{4!}}}{3! \cancel{\color{Cerulean}{4!}}} \\[4pt]&=\frac{210}{6} \\[4pt] &=35\end{aligned}$

Answer:

$35$

Check the menu system of your calculator for a function that calculates this quantity. Look for the notation $_{n} C_{k}$ in the probability subsection.

Exercise $\PageIndex{1}$

Calculate $\left( \begin{array}{l}{8} \\[4pt] {5}\end{array}\right)$.

Answer

$56$

www.youtube.com/v/Rpb8KD1HQGc

Consider the following binomial raised to the $3^{rd}$ power in its expanded form:

$(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$

Compare it to the following calculations,

$\left( \begin{array}{l}{3} \\[4pt] {0}\end{array}\right)=\dfrac{3 !}{0 !(3-0) !}=\frac{3 !}{1 \cdot 3 !}=1$
$\left( \begin{array}{l}{3} \\[4pt] {1}\end{array}\right)=d\frac{3 !}{1 !(3-1) !}=\frac{3 \cdot 2 !}{1 \cdot 2 !}=3$
$\left( \begin{array}{l}{3} \\[4pt] {2}\end{array}\right)=\dfrac{3 !}{2 !(3-2) !}=\frac{3 \cdot 2 !}{2 !}=3$
$\left( \begin{array}{c}{3} \\[4pt] {3}\end{array}\right)=\dfrac{3 !}{3 !(3-3) !}=\frac{3 !}{3 ! 0 !}=1$

Notice that there appears to be a connection between these calculations and the coefficients of the expanded binomial. This observation is generalized in the next section.

Binomial Theorem

Consider expanding $(x+2)^{5}$:

$(x+2)^{5}=(x+2)(x+2)(x+2)(x+2)(x+2)$

One quickly realizes that this is a very tedious calculation involving multiple applications of the distributive property. The binomial theorem²⁸ provides a method of expanding binomials raised to powers without directly multiplying each factor:

$(x+y)^{n}=\left( \begin{array}{c}{n} \\[4pt] {0}\end{array}\right) x^{n} y^{0}+\left( \begin{array}{c}{n} \\[4pt] {1}\end{array}\right) x^{n-1} y^{1}+\left( \begin{array}{c}{n} \\[4pt] {2}\end{array}\right) x^{n-2} y^{2}+\ldots+\left( \begin{array}{c}{n} \\[4pt] {n-1}\end{array}\right) x^{1} y^{n-1}$

More compactly we can write,

$$(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k} \quad \color{Cerulean}{Binomial\:theorem} \nonumber$$

Example $\PageIndex{3}$

Expand using the binomial theorem: $(x + 2)^{5}$.

Solution

Use the binomial theorem where $n = 5$ and $y = 2$.

$(x+2)^{5}=\left( \begin{array}{l}{5} \\[4pt] {0}\end{array}\right) x^{5} 2^{0}+\left( \begin{array}{c}{5} \\[4pt] {1}\end{array}\right) x^{4} 2^{1}+\left( \begin{array}{l}{5} \\[4pt] {2}\end{array}\right) x^{3} 2^{2}+\left( \begin{array}{l}{5} \\[4pt] {3}\end{array}\right) x^{2} 2^{3}+\left( \begin{array}{l}{5} \\[4pt] {4}\end{array}\right) x^{1}2^{4}$

Sometimes it is helpful to identify the pattern that results from applying the binomial theorem. Notice that powers of the variable $x$ start at $5$ and decrease to zero. The powers of the constant term start at $0$ and increase to $5$. The binomial coefficients can be calculated off to the side and are left to the reader as an exercise.

$\begin{aligned}(x+2)^{5} &=\left( \begin{array}{c}{5} \\[4pt] {0}\end{array}\right) x^{5} 2^{0}+\left( \begin{array}{c}{5} \\[4pt] {1}\end{array}\right) x^{4} 2^{1}+\left( \begin{array}{c}{5} \\[4pt] {2}\end{array}\right) x^{3} 2^{2}+\left( \begin{array}{c}{5} \\[4pt] {3}\end{array}\right) x^{2} 2^{3}+\left( \begin{array}{c}{5} \\[4pt] {4}\end{array}\right) x^{1} 2^{4} \\[4pt] &=1 x^{5} \times 1+5 x^{4} \times 2+10 x^{3} \times 4+10 x^{2} \times 8+5 x^{1} \times 16+1 \times 1 \\[4pt] &=x^{5}+10 x^{4}+40 x^{3}+80 x^{2}+80 x+32 \end{aligned}$

Answer

$x^{5}+10 x^{4}+40 x^{3}+80 x^{2}+80 x+32$

The binomial may have negative terms, in which case we will obtain an alternating series.

Example $\PageIndex{4}$

Expand using the binomial theorem: $(u − 2v)^{4}$.

Solution

Use the binomial theorem where $n = 4, x = u$, and $y = −2v$ and then simplify each term.

$\begin{aligned}(u-2 v)^{4} &=\left( \begin{array}{c}{4} \\[4pt] {0}\end{array}\right) u^{4}(-2 v)^{0}+\left( \begin{array}{c}{4} \\[4pt] {1}\end{array}\right) u^{3}(-2 v)^{1}+\left( \begin{array}{c}{4} \\[4pt] {2}\end{array}\right) u^{2}(-2 v)^{2}+\left( \begin{array}{c}{4} \\[4pt] {3}\end{array}\right) u^{1}(-2v)^{3} + \left(\begin{array}{c}4 \\[4pt]4 \end{array} \right)u^{0}(-2v)^{4} \\[4pt] &=1 \times u^{4} \times 1+4 u^{3}(-2 v)+6 u^{2}\left(4 v^{2}\right)+4 u\left(-8 v^{3}\right) + 16v^{4} \\[4pt] &=u^{4}-8 u^{3} v+24 u^{2} v^{2}-32 u v^{3}+16 v^{4} \end{aligned}$

Answer

$u^{4}-8 u^{3} v+24 u^{2} v^{2}-32 u v^{3}+16 v^{4}$

Exercise $\PageIndex{2}$

Expand using the binomial theorem: $\left(a^{2}-3\right)^{4}$

Answer

$a^{8}-12 a^{6}+54 a^{4}-108 a^{2}+81$

www.youtube.com/v/wICbqmoa4T4

Next we study the coefficients of the expansions of $(x + y)^{n}$ starting with $n = 0$:

$$\begin{align*} (x+y)^{0} &=1 \\[4pt] (x+y)^{1}&=x+y \\[4pt] (x+y)^{2}&=x+y \\[4pt] (x+y)^{3}&=x^{2}+2 x y+y^{2} \\[4pt] (x+y)^{3}&=x^{3}+3 x^{2} y+3 x y^{2}+y^{3} \\[4pt] (x+y)^{4}&=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4} \end{align*}$$

Write the coefficients in a triangular array and note that each number below is the sum of the two numbers above it, always leaving a $1$ on either end.

Figure $\PageIndex{1}$

This is Pascal’s triangle²⁹; it provides a quick method for calculating the binomial coefficients. Use this in conjunction with the binomial theorem to streamline the process of expanding binomials raised to powers. For example, to expand $(x − 1)^{6}$ we would need two more rows of Pascal’s triangle,

Figure $\PageIndex{2}$

The binomial coefficients that we need are in blue. Use these numbers and the binomial theorem to quickly expand $(x − 1)^{6}$ as follows:

$\begin{aligned}(x-1)^{6} &=1 x^{6}(-1)^{0}+6 x^{5}(-1)^{1}+15 x^{4}(-1)^{2}+20 x^{3}(-1)^{3}+15 x^{2}(-1)^{4}+6 x(-1)^{5} \\[4pt] &=x^{6}-6 x^{5}+15 x^{4}-20 x^{3}+15 x^{2}-6 x+1 \end{aligned}$

Example $\PageIndex{5}$

Expand using the binomial theorem and Pascal’s triangle: $(2x − 5)^{4}$.

Solution

From Pascal’s triangle we can see that when $n = 4$ the binomial coefficients are $1, 4, 6, 4$, and $1$.Use these numbers and the binomial theorem as follows:

$\begin{aligned}(2 x-5)^{4} &=1(2 x)^{4}(-5)^{0}+4(2 x)^{3}(-5)^{1}+6(2 x)^{2}(-5)^{2}+4(2 x)^{1}(-5)^{3}+(2 x)^{0}(-5)^{4} \\[4pt] &=16 x^{4} \cdot 1+4 \cdot 8 x^{3}(-5)+6 \cdot 4 x^{2} \cdot 25+4 \cdot 2 x(-125)+1 \cdot 625 \\[4pt] &=16 x^{4}-160 x^{3}+600 x^{2}-1,000 x+625 \end{aligned}$

Answer:

$16 x^{4}-160 x^{3}+600 x^{2}-1,000 x+625$

Key Takeaways

• To calculate the factorial of a natural number, multiply that number by all natural numbers less than it: $5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120$. Remember that we have defined $0! = 1$.
• The binomial coefficients are the integers calculated using the formula: $$\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}. \nonumber$$
• The binomial theorem provides a method for expanding binomials raised to powers without directly multiplying each factor: $$(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k}\nonumber$$
• Use Pascal’s triangle to quickly determine the binomial coefficients.

Exercise $\PageIndex{3}$

Evaluate.

1. $6!$
2. $4!$
3. $10!$
4. $9!$
5. $\frac{6 !}{3 !}$
6. $\frac{8 !}{4 !}$
7. $\frac{13 !}{9 !}$
8. $\frac{15 !}{10 !}$
9. $\frac{12 !}{3 ! 7 !}$
10. $\frac{10 !}{2 ! 5 !}$
11. $\frac{n !}{(n-2) !}$
12. $\frac{(n+1) !}{(n-1) !}$
13. (a) $4 !+3 !$ (b) $(4+3) !$
14. (a) $4 !-3 !$ (b) $(4-3) !$

Answer

1. $720$

3. $3,628,800$

5. $120$

7. $17,160$

9. $15,840$

11. $n^{2} − n$

13. a. $30$ b. $5,040$

Exercise $\PageIndex{4}$

Rewrite using factorial notation.

1. $1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7$
2. $1 \times 2 \times 3 \times 4 \times 5$
3. $15 \times 14 \times 13$
4. $10 \times 9 \times 8 \times 7$
5. $13$
6. $8 \times 7$
7. $n(n-1)(n-2)$
8. $1 \times 2 \times 3 \times \cdots \times n \times(n+1)$

Answer

1. $7!$

3. $\frac{15 !}{12 !}$

5. $\frac{13 !}{12 !}$

7. $\frac{n !}{(n-3) !}$

Exercise $\PageIndex{5}$

Calculate the indicated binomial coefficient.

1. $\left( \begin{array}{l}{6} \\[4pt] {4}\end{array}\right)$
2. $\left( \begin{array}{l}{8} \\[4pt] {4}\end{array}\right)$
3. $\left( \begin{array}{l}{7} \\[4pt] {2}\end{array}\right)$
4. $\left( \begin{array}{l}{9} \\[4pt] {5}\end{array}\right)$
5. $\left( \begin{array}{l}{9} \\[4pt] {0}\end{array}\right)$
6. $\left( \begin{array}{l}{13} \\[4pt] {12}\end{array}\right)$
7. $\left( \begin{array}{l}{n} \\[4pt] {0}\end{array}\right)$
8. $\left( \begin{array}{l}{n} \\[4pt] {n}\end{array}\right)$
9. $\left( \begin{array}{l}{n} \\[4pt] {1}\end{array}\right)$
10. $\left( \begin{array}{c}{n} \\[4pt] {n-1}\end{array}\right)$
11. $_{10} C_{8}$
12. $_{5} C_{1}$
13. $_{12} C_{12}$
14. $_{10} C_{5}$
15. $_{n} C_{n-2}$
16. $_{n} C_{n-3}$

Answer

1. $15$

3. $21$

5. $1$

7. $1$

9. $n$

11. $45$

13. $1$

15. $\frac{n^{2}-n}{2}$

Exercise $\PageIndex{6}$

Expand using the binomial theorem.

1. $(4 x-3)^{3}$
2. $(2 x-5)^{3}$
3. $\left(\frac{x}{2}+y\right)^{3}$
4. $\left(x+\frac{1}{y}\right)^{3}$
5. $(x+3)^{4}$
6. $(x+5)^{4}$
7. $(x-4)^{4}$
8. $(x-2)^{4}$
9. $\left(x+\frac{2}{y}\right)^{4}$
10. $\left(\frac{x}{3}-y\right)^{4}$
11. $(x+1)^{5}$
12. $(x-3)^{5}$
13. $(x-2)^{6}$
14. $(x+1)^{6}$
15. $(x-1)^{7}$
16. $(x+1)^{7}$
17. $(5 x-1)^{4}$
18. $(3 x-2)^{4}$
19. $(4 u+v)^{4}$
20. $(3 u-v)^{4}$
21. $(u-5 v)^{5}$
22. $(2 u+3 v)^{5}$
23. $\left(a-b^{2}\right)^{5}$
24. $\left(a^{2}+b^{2}\right)^{4}$
25. $\left(a^{2}+b^{4}\right)^{6}$
26. $\left(a^{5}+b^{2}\right)^{5}$
27. $(x+\sqrt{2})^{3}$
28. $(x-\sqrt{2})^{4}$
29. $(\sqrt{x}-\sqrt{y})^{4}, x, y \geq 0$
30. $(\sqrt{x}+2 \sqrt{y})^{5}, x, y \geq 0$
31. $(x+y)^{7}$
32. $(x+y)^{8}$
33. $(x+y)^{9}$
34. $(x-y)^{7}$
35. $(x-y)^{8}$
36. $(x-y)^{9}$

Answer

1. $64 x^{3}-144 x^{2}+108 x-27$

3. $\frac{x^{3}}{8}+\frac{3 x^{2} y}{4}+\frac{3 x y^{2}}{2}+y^{3}$

5. $x^{4}+12 x^{3}+54 x^{2}+108 x+81$

7. $x^{4}-16 x^{3}+96 x^{2}-256 x+256$

9. $x^{4}+\frac{8 x^{3}}{y}+\frac{24 x^{2}}{y^{2}}+\frac{32 x}{y^{3}}+\frac{16}{y^{4}}$

11. $x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x+1$

13. $x^{6}-12 x^{5}+60 x^{4}-160 x^{3}+240 x^{2}-192 x+64$

15. $x^{7}-7 x^{6}+21 x^{5}-35 x^{4}+35 x^{3}-21 x^{2}+7 x-1$

17. $625 x^{4}-500 x^{3}+150 x^{2}-20 x+1$

19. $256 u^{4}+256 u^{3} v+96 u^{2} v^{2}+16 u v^{3}+v^{4}$

21. $\begin{array}{l}{u^{5}-25 u^{4} v+250 u^{3} v^{2}-1,250 u^{2} v^{3}} {+3,125 u v^{4}-3,125 v^{5}}\end{array}$

23. $a^{5}-5 a^{4} b^{2}+10 a^{3} b^{4}-10 a^{2} b^{6}+5 a b^{8}-b^{10}$

25. $\begin{array}{l}{a^{12}+6 a^{10} b^{4}+15 a^{8} b^{8}+20 a^{6} b^{12}} {+15 a^{4} b^{16}+6 a^{2} b^{20}+b^{24}}\end{array}$

27. $x^{3}+3 \sqrt{2} x^{2}+6 x+2 \sqrt{2}$

29. $x^{2}-4 x \sqrt{x y}+6 x y-4 y \sqrt{x y}+y^{2}$

31. $\begin{array}{l}{x^{7}+7 x^{6} y+21 x^{5} y^{2}+35 x^{4} y^{3}} {+35 x^{3} y^{4}+21 x^{2} y^{5}+7 x y^{6}+y^{7}}\end{array}$

33. $\begin{array}{l}{x^{9}+9 x^{8} y+36 x^{7} y^{2}+84 x^{6} y^{3}+126 x^{5} y^{4}} {+126 x^{4} y^{5}+84 x^{3} y^{6}+36 x^{2} y^{7}+9 x y^{8}+y^{9}}\end{array}$

35. $\begin{array}{l}{x^{8}-8 x^{7} y+28 x^{6} y^{2}-56 x^{5} y^{3}+70 x^{4} y^{4}} {-56 x^{3} y^{5}+28 x^{2} y^{6}-8 x y^{7}+y^{8}}\end{array}$

Exercise $\PageIndex{7}$

1. Determine the factorials of the integers $5, 10, 15, 20$, and $25$. Which grows faster, the common exponential function $a_{n} = 10^{n}$ or the factorial function $a_{n} = n!$? Explain.
2. Research and discuss the history of the binomial theorem.

Answer

1. Answer may vary

Footnotes

²⁵The product of all natural numbers less than or equal to a given natural number, denoted $n!$.

²⁶The factorial of zero is defined to be equal to $1; 0! = 1$.

²⁷An integer that is calculated using the formula: $\left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right)=\frac{n !}{k !(n-k) !}$

²⁸Describes the algebraic expansion of binomials raised to powers: $(x+y)^{n}=\sum_{k=0}^{n} \left( \begin{array}{l}{n} \\[4pt] {k}\end{array}\right) x^{n-k} y^{k}$.

²⁹A triangular array of numbers that correspond to the binomial coefficients.