9.6 Binomial theorem  (Page 2/6)

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• Introduce ${x}^{5},$ and then for each successive term reduce the exponent on $x$ by 1 until ${x}^{0}=1$ is reached.
• Introduce ${y}^{0}=1,$ and then increase the exponent on $y$ by 1 until ${y}^{5}$ is reached.
${x}^{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{4}y,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{3}{y}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}{y}^{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}x{y}^{4},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{y}^{5}$

The next expansion would be

${\left(x+y\right)}^{5}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}.$

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle , shown in [link] .

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3 rd row, flank the ends of the rows with 1’s, and add $1+1$ to find the middle number, 2. In the $n\text{th}$ row, flank 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 coefficients, let us revisit the expansion of the binomials in general form. The binomial theorem

The Binomial Theorem    is a formula that can be used to expand any binomial.

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

Given a binomial, write it in expanded form.

1. Determine the value of $n$ according to the exponent.
2. Evaluate the $k=0$ through $k=n$ using the Binomial Theorem formula.
3. Simplify.

Expanding a binomial

Write in expanded form.

1. $\text{\hspace{0.17em}}{\left(x+y\right)}^{5}\text{\hspace{0.17em}}$
2. $\text{\hspace{0.17em}}{\left(3x-y\right)}^{4}\text{\hspace{0.17em}}$
1. Substitute $n=5$ into the formula. Evaluate the $k=0$ through $k=5$ terms. Simplify.
$\begin{array}{ll}{\left(x+y\right)}^{5}\hfill & =\left(\begin{array}{c}5\\ 0\end{array}\right){x}^{5}{y}^{0}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}{y}^{1}+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right){x}^{1}{y}^{4}+\left(\begin{array}{c}5\\ 5\end{array}\right){x}^{0}{y}^{5}\hfill \\ {\left(x+y\right)}^{5}\hfill & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\hfill \end{array}$
2. 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.
$\begin{array}{ll}{\left(3x-y\right)}^{4}\hfill & =\left(\begin{array}{c}4\\ 0\end{array}\right){\left(3x\right)}^{4}{\left(-y\right)}^{0}+\left(\begin{array}{c}4\\ 1\end{array}\right){\left(3x\right)}^{3}{\left(-y\right)}^{1}+\left(\begin{array}{c}4\\ 2\end{array}\right){\left(3x\right)}^{2}{\left(-y\right)}^{2}+\left(\begin{array}{c}4\\ 3\end{array}\right){\left(3x\right)}^{1}{\left(-y\right)}^{3}+\left(\begin{array}{c}4\\ 4\end{array}\right){\left(3x\right)}^{0}{\left(-y\right)}^{4}\hfill \\ {\left(3x-y\right)}^{4}\hfill & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\hfill \end{array}$

Write in expanded form.

1. ${\left(x-y\right)}^{5}$
2. ${\left(2x+5y\right)}^{3}$
1. ${x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}$
2. $8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}$

Using the binomial theorem to find a single term

Expanding a binomial with a high exponent such as $\text{\hspace{0.17em}}{\left(x+2y\right)}^{16}\text{\hspace{0.17em}}$ 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 find a single specific term.

Note the pattern of coefficients in the expansion of $\text{\hspace{0.17em}}{\left(x+y\right)}^{5}.$

${\left(x+y\right)}^{5}={x}^{5}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right)x{y}^{4}+{y}^{5}$

The second term is $\text{\hspace{0.17em}}\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y.\text{\hspace{0.17em}}$ The third term is $\text{\hspace{0.17em}}\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}.\text{\hspace{0.17em}}$ We can generalize this result.

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

The (r+1)th term of a binomial expansion

The $\text{\hspace{0.17em}}\left(r+1\right)\text{th}\text{\hspace{0.17em}}$ term of the binomial expansion    of $\text{\hspace{0.17em}}{\left(x+y\right)}^{n}\text{\hspace{0.17em}}$ is:

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$

Given a binomial, write a specific term without fully expanding.

1. Determine the value of $n$ according to the exponent.
2. Determine $\left(r+1\right).$
3. Determine $r.$
4. Replace $r$ in the formula for the $\left(r+1\right)\text{th}$ term of the binomial expansion.

Writing a given term of a binomial expansion

Find the tenth term of $\text{\hspace{0.17em}}{\left(x+2y\right)}^{16}\text{\hspace{0.17em}}$ without fully expanding the binomial.

Because we are looking for the tenth term, $\text{\hspace{0.17em}}r+1=10,\text{\hspace{0.17em}}$ we will use $\text{\hspace{0.17em}}r=9$ in our calculations.

$\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}$
$\left(\begin{array}{c}16\\ 9\end{array}\right){x}^{16-9}{\left(2y\right)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}$

Find the sixth term of $\text{\hspace{0.17em}}{\left(3x-y\right)}^{9}\text{\hspace{0.17em}}$ without fully expanding the binomial.

$\text{\hspace{0.17em}}-10,206{x}^{4}{y}^{5}$

Access these online resources for additional instruction and practice with binomial expansion.

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x+2y-z=7
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-1
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AJ
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Atone
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I know this work
salma
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Abhi
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Augustine
how do they get the third part x = (32)5/4
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AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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hmm
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