To expand $(2 + x)^5$, we can use the Binomial Theorem. According to the Binomial Theorem:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case, $a = 2$, $b = x$, and $n = 5$. We calculate each term of the expansion as follows:

$(2 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} x^k$

Now, let's compute each term:

1. $k = 0$: $\binom{5}{0} 2^{5} x^0 = 1 \cdot 32 \cdot 1 = 32$
2. $k = 1$: $\binom{5}{1} 2^{4} x^1 = 5 \cdot 16 \cdot x = 80x$
3. $k = 2$: $\binom{5}{2} 2^{3} x^2 = 10 \cdot 8 \cdot x^2 = 80x^2$
4. $k = 3$: $\binom{5}{3} 2^{2} x^3 = 10 \cdot 4 \cdot x^3 = 40x^3$
5. $k = 4$: $\binom{5}{4} 2^{1} x^4 = 5 \cdot 2 \cdot x^4 = 10x^4$
6. $k = 5$: $\binom{5}{5} 2^{0} x^5 = 1 \cdot 1 \cdot x^5 = x^5$

Now, summing these terms together:

$(2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$

Thus, the expanded form is:

$(2 + x)^5 = x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32$

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