OpenStax-CNX module: m49449
1
Binomial Theorem
∗
OpenStax College
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 4.0
†
Abstract
In this section, you will:
•
Apply the Binomial Theorem.
A polynomial with two terms is called a binomial. We have already learned to multiply binomials and
to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In
this section, we will discuss a shortcut that will allow us to nd (x + y)n without multiplying the binomial
by itself n times.
1 Identifying Binomial Coecients
In Counting Principles1 , we studied combinations. In the shortcut to nding (x + y)n , we will need to use
combinationsto nd
 the coecients that will appear in the expansion of the binomial. In this case, we use
n
the notation   instead of C (n, r) , but it can be calculated in the same way. So
r



The combination 
n
r
n
r

 = C (n, r) =
n!
r! (n − r)!

(1)

 is called a binomial coecient. An example of a binomial coecient is 
C (5, 2) = 10.
A General Note:
If n and rare integers greater than or equal to 0 with n ≥ r, then the
binomial coecient is


∗ Version
n
r

 = C (n, r) =
n!
r! (n − r)!
1.7: Dec 18, 2014 3:59 pm -0600
† http://creativecommons.org/licenses/by/4.0/
1 "Counting Principles" <http://legacy.cnx.org/content/m49448/latest/>
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(2)
5
2

=
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Q&A: Is a binomial coecient always a whole number?
Yes. Just as the number of combinations must always be a whole number, a binomial coecient
will always be a whole number.
Example 1
Finding Binomial Coecients
Find each binomial coecient.


5
a.  
3

b. 

c. 

9

2

9

7
Solution
Use the formula to calculate each binomial coecient. You can also use the nCr function on your
calculator.

n


a. 

b. 

c. 
r

 = C (n, r) =
n!
r! (n − r)!
(3)

5
=
3
5!
3!(5−3)!
=
5·4·3!
3!2!
= 10
9!
2!(9−2)!
=
9·8·7!
2!7!
= 36
9!
7!(9−7)!
=
9·8·7!
7!2!
= 36

9
=
2

9
=
7
Analysis
Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution
for these two parts, you will see that you end up with the same two factorials in the denominator,
but the order is reversed, just as with combinations.


Try It:
n
r


=
n
n−r

(4)

Find each binomial coecient.
Exercise 2
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(Solution on p. 12.)
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
a. 

b. 
3

7

3
11


4
2 Using the Binomial Theorem
When we expand (x + y)n by multiplying, the result is called a binomial expansion, and it includes
binomial coecients. If we wanted to expand (x + y)52 , we might multiply (x + y) by itself fty-two times.
This could take hours! If we examine some simple binomial expansions, we can nd patterns that will lead
us to a shortcut for nding more complicated binomial expansions.
2
(x + y) = x2 + 2xy + y 2
(5)
3
(x + y) = x3 + 3x2 y + 3xy 2 + y 3
4
(x + y) = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4
First, let's examine the exponents. With each successive term, the exponent for x decreases and the exponent
for y increases. The sum of the two exponents is n for each term.
Next, let's examine the coecients. Notice that the coecients increase and then decrease in a symmetrical pattern. The coecients follow a pattern:


n
0
 
,
n
1
 
,
n
2


 , ..., 
n
n

(6)
.
These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

n
(x + y)
=
Pn
k=0
n


= xn + 
k
n
1

 xn−k y k


 xn−1 y + 
n
2


 xn−2 y 2 + ... + 
n
n−1

(7)
 xy n−1 + y n
Another way to see the coecients is to examine the expansion of a binomial in general form, x + y, to
successive powers 1, 2, 3, and 4.
1
(x + y) = x + y
2
(x + y) = x2 + 2xy + y 2
3
(x + y) = x3 + 3x2 y + 3xy 2 + y 3
4
(x + y) = x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4
Can you guess the next expansion for the binomial (x + y)5 ?
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Figure 1
See Figure 1, which illustrates the following:
•
•
•
•
•
There are n + 1 terms in the expansion of (x + y)n .
The degree (or sum of the exponents) for each term is n.
The powers on x begin with n and decrease to 0.
The powers on y begin with 0 and increase to n.
The coecients are symmetric.
To determine the expansion on (x + y)5 , we see n = 5, thus, there will be 5+1 = 6 terms. Each term has a
combined degree of 5. In descending order for powers of x, the pattern is as follows:
• Introduce x5 , and then for each successive term reduce the exponent on x by 1 until x0 = 1 is reached.
• Introduce y 0 = 1, and then increase the exponent on y by 1 until y 5 is reached.
x5 , x4 y, x3 y 2 , x2 y 3 , xy 4 , y 5
(9)
The next expansion would be
5
(x + y) = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5 .
(10)
But where do those coecients come from? The binomial coecients are symmetric. We can see these
coecients in an array known as Pascal's Triangle, shown in Figure 2.
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Figure 2
To generate Pascal's Triangle, we start by writing a 1. In the row below, row 2, we write two 1's. In the
3rd row, ank the ends of the rows with 1's, and add 1 + 1 to nd the middle number, 2. In the nth row,
ank the ends of the row with 1's. Each element in the triangle is the sum of the two elements immediately
above it.
To see the connection between Pascal's Triangle and binomial coecients, let us revisit the expansion of
the binomials in general form.
A General Note:
The Binomial Theorem is a formula that can be used to expand any
binomial.

(x + y)
n
=
Pn
k=0


= xn + 
How To:
n
k
n
1

 xn−k y k


 xn−1 y + 
n
2

 xn−2 y 2 + ... + 
Given a binomial, write it in expanded form.
1.Determine the value of naccording to the exponent.
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
n
n−1

 xy n−1 + y n
(11)
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2.Evaluate the k = 0 through k = n using the Binomial Theorem formula.
3.Simplify.
Example 2
Expanding a Binomial
Write in expanded form.
a. (x + y)5
b. (3x − y)4
Solution
a. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify.
(x + y)
5
=
5
!
x5 y 0 +
5
!
x4 y 1 +
!
x3 y 2 +
2
1
0
5
5
!
x2 y 3 +
5
!
x1 y 4 +
4
3
5
!
5(12)
x0 y 5
(x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5
b. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3x
is in the place that was occupied by x and that − − y is in the place that was occupied by y.
So we substitute them. Simplify.
(3x − y)4 =
4
!
0
4
(3x − y)
(3x)4 (−y)0 +
4
!
(3x)3 (−y)1 +
1
4
!
(3x)2 (−y)2 +
2
= 81x4 − 108x3 y + 54x2 y 2 − 12xy 3 + y 4
Analysis
Notice the alternating signs in part b. This happens because (−y) raised to odd powers is negative,
but (−y) raised to even powers is positive. This will occur whenever the binomial contains a
subtraction sign.
Try It:
Exercise 4
Write in expanded form.
a.(x − y)5
b.(2x + 5y)3
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(Solution on p. 12.)
4
3
!
(3x)1 (−y)3 +
(13)
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3 Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as (x + 2y)16 can be a lengthy process.
Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully
expand a binomial to nd a single specic term.
Note the pattern of coecients in the expansion of (x + y)5 .

5
(x + y) = x5 + 

The second term is 
5
1
5
1


 x4 y + 

 x4 y. The third term is 
n
r

 x3 y 2 + 


5
2
5
3


 x2 y 3 + 
5
4

(14)
 xy 4 + y 5

 x3 y 2 . We can generalize this result.

(15)
 xn−r y r
The (r + 1) th term of the binomial expansion of (x + y)n is:


How To:
2


A General Note:
5
n
r

 xn−r y r
(16)
Given a binomial, write a specic term without fully expanding.
1.Determine the value of n according to the exponent.
2.Determine (r + 1) .
3.Determine r.
4.Replace r in the formula for the (r + 1) th term of the binomial expansion.
Example 3
Writing a Given Term of a Binomial Expansion
Find the tenth term of (x + 2y)16 without fully expanding the binomial.
Solution
Because we are looking for the tenth term, r + 1 = 10, we will use r = 9 in our calculations.


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16
9

 xn−r y r
(17)
 x16−9 (2y)9 = 5,857,280x7 y 9
(18)


n
r

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Try It:
Exercise 6
(Solution on p. 12.)
Find the sixth term of (3x − y) without fully expanding the binomial.
9
Media:
pansion.
Access these online resources for additional instruction and practice with binomial ex-
• The Binomial Theorem2
• Binomial Theorem Example3
4 Key Equations

Binomial Theorem
n
(x + y) =

(r + 1) th term of a binomial expansion

n
r
Pn
k−0

n
k

 xn−k y k

 xn−r y r
Table 1
5 Key Concepts

• 
n
r
 is called a binomial coecient and is equal to C (n, r) . See Example 1.
• The Binomial Theorem allows us to expand binomials without multiplying. See Example 2.
• We can nd a given term of a binomial expansion without fully expanding the binomial. See Example 3.
6 Section Exercises
6.1 Verbal
Exercise 7
What is a binomial coecient, and how it is calculated?
(Solution on p. 12.)
Exercise 8
What role do binomial coecients play in a binomial expansion? Are they restricted to any type
of number?
Exercise 9
What is the Binomial Theorem and what is its use?
Exercise 10
When is it an advantage to use the Binomial Theorem? Explain.
2 http://openstaxcollege.org/l/binomialtheorem
3 http://openstaxcollege.org/l/btexample
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(Solution on p. 12.)
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6.2 Algebraic
For the following exercises, evaluate the binomial coecient.
Exercise

 11
(Solution on p. 12.)
6

2

Exercise

 12
5

3

Exercise

 13
(Solution on p. 12.)
7

4

Exercise

 14
9

7

Exercise

 15
(Solution on p. 12.)
10

9

Exercise

 16
25

11

Exercise

 17
(Solution on p. 12.)
17

6

Exercise

18
200

199

For the following exercises, use the Binomial Theorem to expand each binomial.
Exercise 19
(4a − b)
3
(Solution on p. 12.)
Exercise 20
(5a + 2)
3
Exercise 21
3
(Solution on p. 12.)
(3a + 2b)
Exercise 22
4
(2x + 3y)
Exercise 23
5
(4x + 2y)
Exercise 24
4
(3x − 2y)
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(Solution on p. 12.)
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Exercise 25
(Solution on p. 12.)
5
(4x − 3y)
Exercise526
1
x
+ 3y
Exercise 27
x−1 + 2y −1
(Solution on p. 12.)
4
Exercise
28
√
√ 5
x−
y
For the following exercises, use the Binomial Theorem to write the rst three terms of each binomial.
Exercise 29
(Solution on p. 12.)
17
(a + b)
Exercise 30
(x − 1)
18
Exercise 31
(a − 2b)
(Solution on p. 12.)
15
Exercise 32
(x − 2y)
8
Exercise 33
(3a + b)
(Solution on p. 12.)
20
Exercise 34
7
(2a + 4b)
Exercise 35
x3 −
√
y
(Solution on p. 12.)
8
For the following exercises, nd the indicated term of each binomial without fully expanding the binomial.
Exercise 36
The fourth term of (2x − 3y)4
Exercise 37
The fourth term of (3x − 2y)5
(Solution on p. 12.)
Exercise 38
The third term of (6x − 3y)7
Exercise 39
The eighth term of (7 + 5y)14
(Solution on p. 12.)
Exercise 40
The seventh term of (a + b)11
Exercise 41
(Solution on p. 13.)
The fth term of (x − y)7
Exercise 42
The tenth term of (x − 1)12
Exercise 43
(Solution on p. 13.)
The ninth term of a − 3b2
Exercise 44
The fourth term of x3 −
Exercise 45
The eighth term of
11
1 10
2
(Solution on p. 13.)
y
2
+
2 9
x
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6.3 Graphical
For the following exercises, use the Binomial Theorem to expand the binomial f (x) = (x + 3)4 . Then nd
and graph each indicated sum on one set of axes.
Exercise 46
Find and graph f1 (x) , such that f1 (x) is the rst term of the expansion.
Exercise 47
(Solution on p. 13.)
Find and graph f2 (x) , such that f2 (x) is the sum of the rst two terms of the expansion.
Exercise 48
Find and graph f3 (x) , such that f3 (x) is the sum of the rst three terms of the expansion.
Exercise 49
(Solution on p. 14.)
Find and graph f4 (x) , such that f4 (x) is the sum of the rst four terms of the expansion.
Exercise 50
Find and graph f5 (x) , such that f5 (x) is the sum of the rst ve terms of the expansion.
6.4 Extensions
Exercise 51
(Solution on p. 15.)


n
In the expansion of (5x + 3y)n , each term has the form   an−−k bk , where k successively takes
k




n
7
on the value 0, 1, 2, ..., n.If   =   , what is the corresponding term?
k
2
Exercise 52
In the expansion of (a + b)n , the coecient of an−k bk is the same as the coecient of which other
term?
Exercise 53
(Solution on p. 15.)
Consider the expansion of (x + b)40 . What is the exponent of b in the kth term?
Exercise
 54
 
Find 
+
it.
n
k−1
n
k

(Solution
 onp. 15.)
 and write the answer as a binomial coecient in the form 
Hint: Use the fact that, for any integer p, such that p ≥ 1, p! = p (p − 1)!.
Exercise 55
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k
 . Prove
(Solution on p. 16.)
Which expression cannot be expanded using the Binomial Theorem? Explain.
• x2 − 2x + 1
√
√
8
• ( a + 4 a − 5)
5
• x3 + 2y 2 − z
p 12
• 3x2 − 2y 3
n
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Solutions to Exercises in this Module
Solution to Exercise (p. 2)
a. 35
b. 330
Solution to Exercise (p. 6)
a. x5 − 5x4 y + 10x3 y 2 − 10x2 y 3 + 5xy 4 − y 5
b. 8x3 + 60x2 y + 150xy 2 + 125y 3
Solution to Exercise (p. 8)
− 10, 206x4 y 5
Solution to Exercise (p. 8)

A binomial coecient is an alternative way of denoting the combination C (n, r) . It is dened as 
C (n, r) =
The Binomial Theorem is dened as (x + y)n =

Pn
k=0

n
k
 xn−k y k and can be used to expand any
Solution to Exercise (p. 9)
15
Solution to Exercise (p. 9)
35
Solution to Exercise (p. 9)
10
Solution to Exercise (p. 9)
12,376
Solution to Exercise (p. 9)
64a3 − 48a2 b + 12ab2 − b3
Solution to Exercise (p. 9)
27a3 + 54a2 b + 36ab2 + 8b3
Solution to Exercise (p. 9)
1024x5 + 2560x4 y + 2560x3 y 2 + 1280x2 y 3 + 320xy 4 + 32y 5
Solution to Exercise (p. 9)
1024x5 − 3840x4 y + 5760x3 y 2 − 4320x2 y 3 + 1620xy 4 − 243y 5
Solution to Exercise (p. 10)
8
x3 y
+
=

binomial.
+
r

n!
r!(n−r)! .
Solution to Exercise (p. 8)
1
x4
n
24
x2 y 2
+
32
xy 3
+
16
y4
Solution to Exercise (p. 10)
a17 + 17a16 b + 136a15 b2
Solution to Exercise (p. 10)
a15 − 30a14 b + 420a13 b2
Solution to Exercise (p. 10)
3, 486, 784, 401a20 + 23, 245, 229, 340a19 b + 73, 609, 892, 910a18 b2
Solution to
Exercise (p. 10)
√
x24 − 8x21 y + 28x18 y
Solution to Exercise (p. 10)
−720x2 y 3
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OpenStax-CNX module: m49449
Solution to Exercise (p. 10)
220, 812, 466, 875, 000y 7
Solution to Exercise (p. 10)
35x3 y 4
Solution to Exercise (p. 10)
1, 082, 565a3 b16
Solution to Exercise (p. 10)
1152y 2
x7
Solution to Exercise (p. 11)
f2 (x) = x4 + 12x3
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OpenStax-CNX module: m49449
lution to Exercise (p. 11)
f4 (x) = x4 + 12x3 + 54x2 + 108x
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So-
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lution to Exercise (p. 11)
590, 625x5 y 2
Solution to Exercise (p. 11)
k−1
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So-
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Solution
to Exercise



 (p.
 11)
n

n+1
=
k
k
 

n
n

+

k
k−1
k−1

=
=
=
=
=
=
+
n

 ; Proof:
n!
n!
k!(n−k)! + (k−1)!(n−(k−1))!
n!
n!
k!(n−k)! + (k−1)!(n−k+1)!
(n−k+1)n!
kn!
(n−k+1)k!(n−k)! + k(k−1)!(n−k+1)!
(n−k+1)n!+kn!
k!(n−k+1)!
(n+1)n!
k!((n+1)−k)!
(n+1)!
k!((n+1)−k)!


=
n+1
k

Solution to Exercise (p. 11)
The expression x3 + 2y 2 − z
ten as a binomial.
5
cannot be expanded using the Binomial Theorem because it cannot be rewrit-
Glossary
Denition 1: binomial coecient
the number of ways to choose

r objects from n objects where order does not matter; equivalent
n
to C (n, r) , denoted  
r
Denition 2: binomial expansion
the result of expanding (x + y)n by multiplying
Denition 3: Binomial Theorem
a formula that can be used to expand any binomial
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