To determine the coefficient of $x^2$ in the expansion of $(2 + x)^5$, we can use the binomial theorem. The binomial theorem states:

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

where $\binom{n}{k}$ represents the binomial coefficient.

In this problem:

• $a = 2$,
• $b = x$,
• $n = 5$.

We are interested in the term involving $x^2$. For this, $k = 2$, since the exponent of $x$ is $k$.

The corresponding term in the expansion is:

$\binom{5}{2} \cdot 2^{5-2} \cdot x^2.$

Calculate the coefficient:

$\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10,$

and

$2^{5-2} = 2^3 = 8.$

Therefore, the coefficient of $x^2$ is:

$10 \times 8 = 80.$

So, the coefficient of $x^2$ in the expansion of $(2 + x)^5$ is 80.

Would you like more details or have any other questions? Here are five related questions to explore:

1. What is the coefficient of $x^3$ in $(2 + x)^5$?
2. How do you expand $(a + b)^n$ using the binomial theorem?
3. What is the general form of a binomial expansion?
4. Can you determine the coefficient of $x^4$ in $(3 + x)^6$?
5. How would the expansion change if $x$ had a negative coefficient?

Tip: In binomial expansions, always identify the value of $k$ corresponding to the power of $x$ you need to determine the coefficient.