# 13.6 Binomial theorem

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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 find $\text{\hspace{0.17em}}{\left(x+y\right)}^{n}\text{\hspace{0.17em}}$ without multiplying the binomial by itself $n$ times.

## Identifying binomial coefficients

In Counting Principles , we studied combinations . In the shortcut to finding $\text{\hspace{0.17em}}{\left(x+y\right)}^{n},\text{\hspace{0.17em}}$ we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)\text{\hspace{0.17em}}$ instead of $C\left(n,r\right),$ but it can be calculated in the same way. So

$\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\text{\hspace{0.17em}}$

The combination $\text{\hspace{0.17em}}\left(\begin{array}{c}n\\ r\end{array}\right)\text{\hspace{0.17em}}$ is called a binomial coefficient . An example of a binomial coefficient is $\text{\hspace{0.17em}}\left(\begin{array}{c}5\\ 2\end{array}\right)=C\left(5,2\right)=10.\text{\hspace{0.17em}}$

## Binomial coefficients

If $n$ and $r$ are integers greater than or equal to 0 with $n\ge r,$ then the binomial coefficient    is

$\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$

Is a binomial coefficient always a whole number?

Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.

## Finding binomial coefficients

Find each binomial coefficient.

1. $\left(\begin{array}{c}5\\ 3\end{array}\right)$
2. $\left(\begin{array}{c}9\\ 2\end{array}\right)$
3. $\left(\begin{array}{c}9\\ 7\end{array}\right)$

Use the formula to calculate each binomial coefficient. You can also use the ${n}_{}{C}_{r}$ function on your calculator.

$\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$
1. $\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{5!}{3!\left(5-3\right)!}=\frac{5\cdot 4\cdot 3!}{3!2!}=10$
2. $\left(\begin{array}{c}9\\ 2\end{array}\right)=\frac{9!}{2!\left(9-2\right)!}=\frac{9\cdot 8\cdot 7!}{2!7!}=36$
3. $\left(\begin{array}{c}9\\ 7\end{array}\right)=\frac{9!}{7!\left(9-7\right)!}=\frac{9\cdot 8\cdot 7!}{7!2!}=36$

Find each binomial coefficient.

1. $\text{\hspace{0.17em}}\left(\begin{array}{c}7\\ 3\end{array}\right)\text{\hspace{0.17em}}$
2. $\text{\hspace{0.17em}}\left(\begin{array}{c}11\\ 4\end{array}\right)\text{\hspace{0.17em}}$

1. 35
2. 330

## Using the binomial theorem

When we expand ${\left(x+y\right)}^{n}$ by multiplying, the result is called a binomial expansion    , and it includes binomial coefficients. If we wanted to expand ${\left(x+y\right)}^{52},$ we might multiply $\left(x+y\right)$ by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

$\begin{array}{l}{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2}\hfill \\ {\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\hfill \\ {\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\hfill \end{array}$

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 coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

$\left(\begin{array}{c}n\\ 0\end{array}\right),\left(\begin{array}{c}n\\ 1\end{array}\right),\left(\begin{array}{c}n\\ 2\end{array}\right),...,\left(\begin{array}{c}n\\ n\end{array}\right).$

These patterns lead us to the Binomial Theorem , which can be used to expand any binomial.

$\begin{array}{ll}{\left(x+y\right)}^{n}\hfill & =\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}\hfill \\ \hfill & ={x}^{n}+\left(\begin{array}{c}n\\ 1\end{array}\right){x}^{n-1}y+\left(\begin{array}{c}n\\ 2\end{array}\right){x}^{n-2}{y}^{2}+...+\left(\begin{array}{c}n\\ n-1\end{array}\right)x{y}^{n-1}+{y}^{n}\hfill \end{array}$

Another way to see the coefficients is to examine the expansion of a binomial in general form, $\text{\hspace{0.17em}}x+y,\text{\hspace{0.17em}}$ to successive powers 1, 2, 3, and 4.

$\begin{array}{l}{\left(x+y\right)}^{1}=x+y\hfill \\ {\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2}\hfill \\ {\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\hfill \\ {\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\hfill \end{array}$

Can you guess the next expansion for the binomial $\text{\hspace{0.17em}}{\left(x+y\right)}^{5}?\text{\hspace{0.17em}}$

See [link] , which illustrates the following:

• There are $n+1$ terms in the expansion of ${\left(x+y\right)}^{n}.$
• 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 coefficients are symmetric.

To determine the expansion on ${\left(x+y\right)}^{5},$ 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:

#### Questions & Answers

if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
Martin Reply
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
Yanah Reply
hi
John
hi
Grace
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John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
Umesh Reply
acha se dhek ke bata sin theta ke value
Ajay
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Ajay
I want to know trigonometry but I can't understand it anyone who can help
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Yh
Idowu
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Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
Sudip Reply
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
Sebit Reply
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
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Marty Reply
I want to know partial fraction Decomposition.
Adama Reply
classes of function in mathematics
Yazidu Reply
divide y2_8y2+5y2/y2
Sumanth Reply
wish i knew calculus to understand what's going on 🙂
Dashawn Reply
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
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Dashawn
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Dashawn
what's the derivative of 4x^6
Axmed Reply
24x^5
James
10x
Axmed
24X^5
Taieb
Thanks for this helpfull app
Axmed Reply
secA+tanA=2√5,sinA=?
richa Reply
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
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NAVJIT Reply
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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