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1

Solved example of binomial theorem

$\left(x+3\right)^5$
2

We can expand the expression $\left(x+3\right)^5$ using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer $n$. The formula is as follows: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$
The number of terms resulting from the expansion always equals $n + 1$. The coefficients $\left(\begin{matrix}n\\k\end{matrix}\right)$ are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of $a$ decreases, from $n$ to $0$, while the exponent of $b$ increases, from $0$ to $n$. If one of the binomial terms is negative, the positive and negative signs alternate.

$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}3^{0}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
3

Calculate the power $3^{0}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}3^{1}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
4

Calculate the power $3^{1}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}3^{2}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
5

Calculate the power $3^{2}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}3^{3}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
6

Calculate the power $3^{3}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}3^{4}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
7

Calculate the power $3^{4}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}3^{5}$
8

Calculate the power $3^{5}$

$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 1x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
9

Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
10

Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 9x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
11

Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 27x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
12

Calculate the binomial coefficient $\left(\begin{matrix}5\\3\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 81x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
13

Calculate the binomial coefficient $\left(\begin{matrix}5\\4\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 243x^{0}$
14

Calculate the binomial coefficient $\left(\begin{matrix}5\\5\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$

$1x^{5}\frac{5!}{\left(0!\right)\left(5+0\right)!}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
15

$x+0=x$, where $x$ is any expression

$1x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x^{1}\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
16

Any expression to the power of $1$ is equal to that same expression

$1x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
17

Any expression multiplied by $1$ is equal to itself

$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+243x^{0}\frac{5!}{\left(5!\right)\left(5-5\right)!}$
18

Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$

$x^{5}\frac{5!}{\left(0!\right)\left(5!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+1\cdot 243\left(\frac{5!}{\left(5!\right)\left(5-5\right)!}\right)$
19

Simplify the fraction $\frac{5!}{\left(0!\right)\left(5!\right)}$ by $5!$

$x^{5}\frac{1}{1\left(0!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+1\cdot 243\left(\frac{5!}{\left(5!\right)\left(5-5\right)!}\right)$
20

Simplify the fraction $\frac{5!}{\left(5!\right)\left(5-5\right)!}$ by $5!$

$x^{5}\frac{1}{1\left(0!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+1\cdot 243\left(\frac{1}{1\left(5-5\right)!}\right)$
21

Multiply the fraction and term

$\frac{x^{5}}{1\left(0!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+1\cdot 243\left(\frac{1}{1\left(5-5\right)!}\right)$
22

Multiply the fraction and term

$\frac{x^{5}}{1\left(0!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+243\left(\frac{1}{1\left(5-5\right)!}\right)$
23

Multiply the fraction and term

$\frac{x^{5}}{1\left(0!\right)}+3x^{4}\frac{5!}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
24

Multiplying the fraction by $x^{4}$

$\frac{x^{5}}{1\left(0!\right)}+3\left(\frac{\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}\right)+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
25

Multiplying the fraction by $3$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+9x^{3}\frac{5!}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
26

Multiplying the fraction by $x^{3}$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+9\left(\frac{\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}\right)+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
27

Multiplying the fraction by $9$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+27x^{2}\frac{5!}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
28

Multiplying the fraction by $x^{2}$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+27\left(\frac{\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}\right)+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
29

Multiplying the fraction by $27$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}+81x\frac{5!}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
30

Multiplying the fraction by $x$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}+81\left(\frac{x\left(5!\right)}{\left(4!\right)\left(5-4\right)!}\right)+\frac{243}{1\left(5-5\right)!}$
31

Multiplying the fraction by $81$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(5-1\right)!}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}+\frac{81x\left(5!\right)}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
32

Subtract the values $5$ and $-1$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(5-2\right)!}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}+\frac{81x\left(5!\right)}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
33

Subtract the values $5$ and $-2$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(5-3\right)!}+\frac{81x\left(5!\right)}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
34

Subtract the values $5$ and $-3$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(5-4\right)!}+\frac{243}{1\left(5-5\right)!}$
35

Subtract the values $5$ and $-4$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(5-5\right)!}$
36

Subtract the values $5$ and $-5$

$\frac{x^{5}}{1\left(0!\right)}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
37

The factorial of $0$ is $1$

$\frac{x^{5}}{1\cdot 1}+\frac{3\left(5!\right)\left(x^{4}\right)}{\left(1!\right)\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
38

The factorial of $1$ is $1$

$\frac{x^{5}}{1\cdot 1}+\frac{3\left(5!\right)\left(x^{4}\right)}{1\left(4!\right)}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
39

The factorial of $4$ is $24$

$\frac{x^{5}}{1\cdot 1}+\frac{3\left(5!\right)\left(x^{4}\right)}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
40

The factorial of $5$ is $120$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{\left(2!\right)\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
41

The factorial of $2$ is $2$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\left(3!\right)}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
42

The factorial of $3$ is $6$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{9\left(5!\right)\left(x^{3}\right)}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
43

The factorial of $5$ is $120$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{\left(3!\right)\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
44

The factorial of $3$ is $6$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\left(2!\right)}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
45

The factorial of $2$ is $2$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{27\left(5!\right)\left(x^{2}\right)}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
46

The factorial of $5$ is $120$

$\frac{x^{5}}{1\cdot 1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
47

Any expression multiplied by $1$ is equal to itself

$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{1\cdot 24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
48

Any expression multiplied by $1$ is equal to itself

$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{1\left(0!\right)}$
49

Any expression multiplied by $1$ is equal to itself

$\frac{x^{5}}{1}+\frac{120\cdot 3x^{4}}{24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
50

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+\frac{120\cdot 3x^{4}}{24}+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
51

Take $\frac{120}{24}$ out of the fraction

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{120\cdot 9x^{3}}{2\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
52

Take $\frac{120}{2}$ out of the fraction

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{9x^{3}\mathrm{div}\left(120,2\right)}{1\cdot 6}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
53

Take $\frac{9}{6}$ out of the fraction

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1\cdot 1}+\frac{120\cdot 27x^{2}}{6\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
54

Take $\frac{120}{6}$ out of the fraction

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1\cdot 1}+\frac{27x^{2}\mathrm{div}\left(120,6\right)}{1\cdot 2}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
55

Take $\frac{27}{2}$ out of the fraction

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1\cdot 1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1\cdot 1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
56

Multiply $1$ times $1$

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1\cdot 1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
57

Multiply $1$ times $1$

$x^{5}+3x^{4}\mathrm{div}\left(120,24\right)+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
58

Divide $120$ by $24$

$x^{5}+5\cdot 3x^{4}+\frac{x^{3}\mathrm{div}\left(9,6\right)\mathrm{div}\left(120,2\right)}{1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
59

Divide $9$ by $6$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}x^{3}\mathrm{div}\left(120,2\right)}{1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
60

Divide $120$ by $2$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{x^{2}\mathrm{div}\left(27,2\right)\mathrm{div}\left(120,6\right)}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
61

Divide $27$ by $2$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}x^{2}\mathrm{div}\left(120,6\right)}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
62

Divide $120$ by $6$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{81x\left(5!\right)}{\left(4!\right)\left(1!\right)}+\frac{243}{0!}$
63

The factorial of $4$ is $24$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{81x\left(5!\right)}{24\left(1!\right)}+\frac{243}{0!}$
64

The factorial of $1$ is $1$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{81x\left(5!\right)}{24\cdot 1}+\frac{243}{0!}$
65

The factorial of $5$ is $120$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{120\cdot 81x}{24\cdot 1}+\frac{243}{0!}$
66

The factorial of $0$ is $1$

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{120\cdot 81x}{24\cdot 1}+\frac{243}{1}$
67

Any expression multiplied by $1$ is equal to itself

$x^{5}+5\cdot 3x^{4}+\frac{\frac{3}{2}\cdot 60x^{3}}{1}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{120\cdot 81x}{24}+\frac{243}{1}$
68

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+5\cdot 3x^{4}+\frac{3}{2}\cdot 60x^{3}+\frac{\frac{27}{2}\cdot 20x^{2}}{1}+\frac{120\cdot 81x}{24}+\frac{243}{1}$
69

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+5\cdot 3x^{4}+\frac{3}{2}\cdot 60x^{3}+\frac{27}{2}\cdot 20x^{2}+\frac{120\cdot 81x}{24}+\frac{243}{1}$
70

Any expression divided by one ($1$) is equal to that same expression

$x^{5}+5\cdot 3x^{4}+\frac{3}{2}\cdot 60x^{3}+\frac{27}{2}\cdot 20x^{2}+\frac{120\cdot 81x}{24}+243$
71

Take $\frac{120}{24}$ out of the fraction

$x^{5}+5\cdot 3x^{4}+\frac{3}{2}\cdot 60x^{3}+\frac{27}{2}\cdot 20x^{2}+81x\mathrm{div}\left(120,24\right)+243$
72

Multiply $5$ times $3$

$x^{5}+15x^{4}+\frac{3}{2}\cdot 60x^{3}+\frac{27}{2}\cdot 20x^{2}+81x\mathrm{div}\left(120,24\right)+243$
73

Multiply $\frac{3}{2}$ times $60$

$x^{5}+15x^{4}+90x^{3}+\frac{27}{2}\cdot 20x^{2}+81x\mathrm{div}\left(120,24\right)+243$
74

Multiply $\frac{27}{2}$ times $20$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+81x\mathrm{div}\left(120,24\right)+243$
75

Divide $120$ by $24$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+5\cdot 81x+243$
76

Multiply $5$ times $81$

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

Final Answer

$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$

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