Answer
$x^{5}+10x^{4}+40x^{3}+80x^{2}+80x+32$
Work Step by Step
The Binomial Theorem$:$
$(a+b)^{n}=\left(\begin{array}{l}
n\\
0
\end{array}\right)a^{n}+\left(\begin{array}{l}
n\\
1
\end{array}\right)a^{n-1}b+\left(\begin{array}{l}
n\\
2
\end{array}\right)a^{n-2}b^{2}+\cdots+\left(\begin{array}{l}
n\\
n
\end{array}\right)b^{n}$
----------------
$\left(\begin{array}{l}
5\\
0
\end{array}\right)x^{5}\cdot 2^{0}=1\cdot x^{5}\cdot 1=x^{5}$
$\left(\begin{array}{l}
5\\
1
\end{array}\right)x^{4}\cdot 2^{1}=5\cdot x^{4}\cdot 2=10x^{4}$
$\displaystyle \left(\begin{array}{l}
5\\
2
\end{array}\right)x^{3}\cdot 2^{2}=\frac{5(4)}{1(2)}\cdot x^{3}\cdot 2^{2}=40x^{3}$
$\displaystyle \left(\begin{array}{l}
5\\
3
\end{array}\right)x^{2}\cdot 2^{3}=\frac{5(4)(3)}{1(2)(3)}\cdot x^{2}\cdot 2^{3}=80x^{2}$
$\displaystyle \left(\begin{array}{l}
5\\
4
\end{array}\right)x^{1}\cdot 2^{4}=\frac{5(4)(3)(2)}{1(2)(3)(4)}\cdot x^{1}\cdot 2^{4}=80x$
$\left(\begin{array}{l}
5\\
5
\end{array}\right)x^{0}\cdot 2^{5}=1\cdot x^{0}\cdot 2^{5}=32$
$(x+2)^{5}=x^{5}+10x^{4}+40x^{3}+80x^{2}+80x+32$
