1
Here, we show you a step-by-step solved example of binomial theorem. This solution was automatically generated by our smart calculator:
$\left(x+3\right)^5$
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.
$\left(\begin{matrix}5\\0\end{matrix}\right)\cdot 3^{0}x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3^{1}x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
3
Calculate the power $3^{0}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+\left(\begin{matrix}5\\1\end{matrix}\right)\cdot 3^{1}x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
4
Calculate the power $3^{1}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+\left(\begin{matrix}5\\2\end{matrix}\right)\cdot 3^{2}x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
5
Calculate the power $3^{2}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+\left(\begin{matrix}5\\3\end{matrix}\right)\cdot 3^{3}x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
6
Calculate the power $3^{3}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+\left(\begin{matrix}5\\4\end{matrix}\right)\cdot 3^{4}x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
7
Calculate the power $3^{4}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+81\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}+\left(\begin{matrix}5\\5\end{matrix}\right)\cdot 3^{5}x^{0}$
8
Calculate the power $3^{5}$
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+81\left(\begin{matrix}5\\4\end{matrix}\right)x^{1}+243\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}$
9
Any expression to the power of $1$ is equal to that same expression
$1\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+81\left(\begin{matrix}5\\4\end{matrix}\right)x+243\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}$
10
Any expression multiplied by $1$ is equal to itself
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+81\left(\begin{matrix}5\\4\end{matrix}\right)x+243\left(\begin{matrix}5\\5\end{matrix}\right)x^{0}$
11
Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$
$\left(\begin{matrix}5\\0\end{matrix}\right)x^{5}+3\left(\begin{matrix}5\\1\end{matrix}\right)x^{4}+9\left(\begin{matrix}5\\2\end{matrix}\right)x^{3}+27\left(\begin{matrix}5\\3\end{matrix}\right)x^{2}+81\left(\begin{matrix}5\\4\end{matrix}\right)x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
12
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
13
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
14
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
15
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
16
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
17
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
18
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
19
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
20
Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$3\frac{5!}{\left(1!\right)\left(5-1\right)!}x^{4}$
21
The factorial of $1$ is $1$
$3\frac{5!}{1\cdot 1}x^{4}$
22
The factorial of $5$ is $120$
$3\frac{120}{1\cdot 1}x^{4}$
23
Any expression multiplied by $1$ is equal to itself
$120\cdot 3x^{4}$
24
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
25
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
26
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
27
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
28
Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$3\frac{5!}{\left(1!\right)\left(5-1\right)!}x^{4}$
29
The factorial of $1$ is $1$
$3\frac{5!}{1\cdot 1}x^{4}$
30
The factorial of $5$ is $120$
$3\frac{120}{1\cdot 1}x^{4}$
31
Any expression multiplied by $1$ is equal to itself
$120\cdot 3x^{4}$
32
Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$9\frac{5!}{\left(2!\right)\left(5-2\right)!}x^{3}$
33
The factorial of $2$ is $2$
$9\frac{5!}{2\cdot 1}x^{3}$
34
The factorial of $5$ is $120$
$9\frac{120}{2\cdot 1}x^{3}$
35
Any expression multiplied by $1$ is equal to itself
$9\frac{120}{2}x^{3}$
36
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
37
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
38
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
39
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
40
Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$3\frac{5!}{\left(1!\right)\left(5-1\right)!}x^{4}$
41
The factorial of $1$ is $1$
$3\frac{5!}{1\cdot 1}x^{4}$
42
The factorial of $5$ is $120$
$3\frac{120}{1\cdot 1}x^{4}$
43
Any expression multiplied by $1$ is equal to itself
$120\cdot 3x^{4}$
44
Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$9\frac{5!}{\left(2!\right)\left(5-2\right)!}x^{3}$
45
The factorial of $2$ is $2$
$9\frac{5!}{2\cdot 1}x^{3}$
46
The factorial of $5$ is $120$
$9\frac{120}{2\cdot 1}x^{3}$
47
Any expression multiplied by $1$ is equal to itself
$9\frac{120}{2}x^{3}$
48
Calculate the binomial coefficient $\left(\begin{matrix}5\\3\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$27\frac{5!}{\left(3!\right)\left(5-3\right)!}x^{2}$
49
The factorial of $3$ is $6$
$27\frac{5!}{6\cdot 1}x^{2}$
50
The factorial of $5$ is $120$
$27\frac{120}{6\cdot 1}x^{2}$
51
Any expression multiplied by $1$ is equal to itself
$27\frac{120}{6}x^{2}$
52
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
53
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
54
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
55
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
56
Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$3\frac{5!}{\left(1!\right)\left(5-1\right)!}x^{4}$
57
The factorial of $1$ is $1$
$3\frac{5!}{1\cdot 1}x^{4}$
58
The factorial of $5$ is $120$
$3\frac{120}{1\cdot 1}x^{4}$
59
Any expression multiplied by $1$ is equal to itself
$120\cdot 3x^{4}$
60
Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$9\frac{5!}{\left(2!\right)\left(5-2\right)!}x^{3}$
61
The factorial of $2$ is $2$
$9\frac{5!}{2\cdot 1}x^{3}$
62
The factorial of $5$ is $120$
$9\frac{120}{2\cdot 1}x^{3}$
63
Any expression multiplied by $1$ is equal to itself
$9\frac{120}{2}x^{3}$
64
Calculate the binomial coefficient $\left(\begin{matrix}5\\3\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$27\frac{5!}{\left(3!\right)\left(5-3\right)!}x^{2}$
65
The factorial of $3$ is $6$
$27\frac{5!}{6\cdot 1}x^{2}$
66
The factorial of $5$ is $120$
$27\frac{120}{6\cdot 1}x^{2}$
67
Any expression multiplied by $1$ is equal to itself
$27\frac{120}{6}x^{2}$
68
Calculate the binomial coefficient $\left(\begin{matrix}5\\4\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$81\frac{5!}{\left(4!\right)\left(5-4\right)!}x$
69
The factorial of $4$ is $24$
$81\frac{5!}{24\cdot 1}x$
70
The factorial of $5$ is $120$
$81\frac{120}{24\cdot 1}x$
71
Any expression multiplied by $1$ is equal to itself
$81\frac{120}{24}x$
72
Subtract the values $5$ and $-1$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(5-2\right)!}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(5-3\right)!}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
73
Subtract the values $5$ and $-2$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(5-3\right)!}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
74
Subtract the values $5$ and $-3$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(5-4\right)!}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
75
Subtract the values $5$ and $-4$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
76
Add the values $5$ and $0$
$\frac{5!}{\left(0!\right)\left(5!\right)}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
77
Simplify the fraction $\frac{5!}{\left(0!\right)\left(5!\right)}$ by $5!$
$\frac{1}{0!}x^{5}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
78
Multiply the fraction by the term
$\frac{1x^{5}}{0!}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
79
Any expression multiplied by $1$ is equal to itself
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+243\left(\begin{matrix}5\\5\end{matrix}\right)$
80
Calculate the binomial coefficient $\left(\begin{matrix}5\\0\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$\frac{5!}{\left(0!\right)\left(5+0\right)!}x^{5}$
81
The factorial of $0$ is $1$
$\frac{5!}{1\cdot 1}x^{5}$
82
The factorial of $5$ is $120$
$\frac{120}{1\cdot 1}x^{5}$
83
Any expression multiplied by $1$ is equal to itself
$120x^{5}$
84
Calculate the binomial coefficient $\left(\begin{matrix}5\\1\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$3\frac{5!}{\left(1!\right)\left(5-1\right)!}x^{4}$
85
The factorial of $1$ is $1$
$3\frac{5!}{1\cdot 1}x^{4}$
86
The factorial of $5$ is $120$
$3\frac{120}{1\cdot 1}x^{4}$
87
Any expression multiplied by $1$ is equal to itself
$120\cdot 3x^{4}$
88
Calculate the binomial coefficient $\left(\begin{matrix}5\\2\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$9\frac{5!}{\left(2!\right)\left(5-2\right)!}x^{3}$
89
The factorial of $2$ is $2$
$9\frac{5!}{2\cdot 1}x^{3}$
90
The factorial of $5$ is $120$
$9\frac{120}{2\cdot 1}x^{3}$
91
Any expression multiplied by $1$ is equal to itself
$9\frac{120}{2}x^{3}$
92
Calculate the binomial coefficient $\left(\begin{matrix}5\\3\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$27\frac{5!}{\left(3!\right)\left(5-3\right)!}x^{2}$
93
The factorial of $3$ is $6$
$27\frac{5!}{6\cdot 1}x^{2}$
94
The factorial of $5$ is $120$
$27\frac{120}{6\cdot 1}x^{2}$
95
Any expression multiplied by $1$ is equal to itself
$27\frac{120}{6}x^{2}$
96
Calculate the binomial coefficient $\left(\begin{matrix}5\\4\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$81\frac{5!}{\left(4!\right)\left(5-4\right)!}x$
97
The factorial of $4$ is $24$
$81\frac{5!}{24\cdot 1}x$
98
The factorial of $5$ is $120$
$81\frac{120}{24\cdot 1}x$
99
Any expression multiplied by $1$ is equal to itself
$81\frac{120}{24}x$
100
Calculate the binomial coefficient $\left(\begin{matrix}5\\5\end{matrix}\right)$ applying the formula: $\left(\begin{matrix}n\\k\end{matrix}\right)=\frac{n!}{k!(n-k)!}$
$243\left(\frac{5!}{\left(5!\right)\left(5-5\right)!}\right)$
101
Simplify the fraction $\frac{5!}{\left(5!\right)\left(5-5\right)!}$ by $5!$
$243\left(\frac{1}{\left(5-5\right)!}\right)$
102
Subtract the values $5$ and $-5$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243\left(5!\right)}{\left(5!\right)\left(0!\right)}$
103
Simplify the fraction $\frac{243\left(5!\right)}{\left(5!\right)\left(0!\right)}$ by $5!$
$\frac{x^{5}}{0!}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
104
The factorial of $0$ is $1$
$\frac{x^{5}}{1}+\frac{3\left(5!\right)}{\left(1!\right)\left(4!\right)}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
105
The factorial of $1$ is $1$
$\frac{x^{5}}{1}+\frac{3\left(5!\right)}{1\cdot 24}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
106
The factorial of $5$ is $120$
$\frac{x^{5}}{1}+\frac{3\cdot 120}{1\cdot 24}x^{4}+\frac{9\left(5!\right)}{\left(2!\right)\left(3!\right)}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
107
The factorial of $2$ is $2$
$\frac{x^{5}}{1}+\frac{3\cdot 120}{1\cdot 24}x^{4}+\frac{9\left(5!\right)}{2\cdot 6}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
108
The factorial of $5$ is $120$
$\frac{x^{5}}{1}+\frac{3\cdot 120}{1\cdot 24}x^{4}+\frac{9\cdot 120}{2\cdot 6}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
109
Multiply $1$ times $24$
$\frac{x^{5}}{1}+\frac{3\cdot 120}{24}x^{4}+\frac{9\cdot 120}{2\cdot 6}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
110
Multiply $3$ times $120$
$\frac{x^{5}}{1}+\frac{360}{24}x^{4}+\frac{9\cdot 120}{2\cdot 6}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
111
Multiply $2$ times $6$
$\frac{x^{5}}{1}+\frac{360}{24}x^{4}+\frac{9\cdot 120}{12}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
112
Multiply $9$ times $120$
$\frac{x^{5}}{1}+\frac{360}{24}x^{4}+\frac{1080}{12}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
$\frac{x^{5}}{1}+15x^{4}+\frac{1080}{12}x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
114
Divide $1080$ by $12$
$\frac{x^{5}}{1}+15x^{4}+90x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
115
Any expression divided by one ($1$) is equal to that same expression
$x^{5}+15x^{4}+90x^{3}+\frac{27\left(5!\right)}{\left(3!\right)\left(2!\right)}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
116
The factorial of $3$ is $6$
$x^{5}+15x^{4}+90x^{3}+\frac{27\left(5!\right)}{6\cdot 2}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
117
The factorial of $5$ is $120$
$x^{5}+15x^{4}+90x^{3}+\frac{27\cdot 120}{6\cdot 2}x^{2}+\frac{81\left(5!\right)}{\left(4!\right)\left(1!\right)}x+\frac{243}{0!}$
118
The factorial of $4$ is $24$
$x^{5}+15x^{4}+90x^{3}+\frac{27\cdot 120}{6\cdot 2}x^{2}+\frac{81\left(5!\right)}{24\cdot 1}x+\frac{243}{0!}$
119
The factorial of $5$ is $120$
$x^{5}+15x^{4}+90x^{3}+\frac{27\cdot 120}{6\cdot 2}x^{2}+\frac{81\cdot 120}{24\cdot 1}x+\frac{243}{0!}$
120
The factorial of $0$ is $1$
$x^{5}+15x^{4}+90x^{3}+\frac{27\cdot 120}{6\cdot 2}x^{2}+\frac{81\cdot 120}{24\cdot 1}x+\frac{243}{1}$
121
Multiply $6$ times $2$
$x^{5}+15x^{4}+90x^{3}+\frac{27\cdot 120}{12}x^{2}+\frac{81\cdot 120}{24\cdot 1}x+\frac{243}{1}$
122
Multiply $27$ times $120$
$x^{5}+15x^{4}+90x^{3}+\frac{3240}{12}x^{2}+\frac{81\cdot 120}{24\cdot 1}x+\frac{243}{1}$
123
Multiply $24$ times $1$
$x^{5}+15x^{4}+90x^{3}+\frac{3240}{12}x^{2}+\frac{81\cdot 120}{24}x+\frac{243}{1}$
124
Multiply $81$ times $120$
$x^{5}+15x^{4}+90x^{3}+\frac{3240}{12}x^{2}+\frac{9720}{24}x+\frac{243}{1}$
125
Divide $3240$ by $12$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+\frac{9720}{24}x+\frac{243}{1}$
126
Divide $9720$ by $24$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+\frac{243}{1}$
$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$
Final answer to the problem
$x^{5}+15x^{4}+90x^{3}+270x^{2}+405x+243$