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:

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
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Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
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Miiro
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AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
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Cliff
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helo
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