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