To find the coefficient of [tex]\( x^2 \)[/tex] in the expansion of [tex]\( (x+2)^3 \)[/tex], let's start by expanding the expression:
[tex]\[
(x + 2)^3
\][/tex]
We can use the binomial theorem to expand this expression. The binomial theorem states that:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\( a = x \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( n = 3 \)[/tex]. Plugging these values into the binomial expansion, we get:
[tex]\[
(x + 2)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} \cdot 2^k
\][/tex]
Let's break this down term by term:
1. For [tex]\( k = 0 \)[/tex]:
[tex]\[
\binom{3}{0} x^{3-0} \cdot 2^0 = 1 \cdot x^3 \cdot 1 = x^3
\][/tex]
2. For [tex]\( k = 1 \)[/tex]:
[tex]\[
\binom{3}{1} x^{3-1} \cdot 2^1 = 3 \cdot x^2 \cdot 2 = 6x^2
\][/tex]
3. For [tex]\( k = 2 \)[/tex]:
[tex]\[
\binom{3}{2} x^{3-2} \cdot 2^2 = 3 \cdot x \cdot 4 = 12x
\][/tex]
4. For [tex]\( k = 3 \)[/tex]:
[tex]\[
\binom{3}{3} x^{3-3} \cdot 2^3 = 1 \cdot 1 \cdot 8 = 8
\][/tex]
Now, combining all the terms, we get the expanded form:
[tex]\[
(x + 2)^3 = x^3 + 6x^2 + 12x + 8
\][/tex]
From this expanded form, we can see that the coefficient of [tex]\( x^2 \)[/tex] is [tex]\( 6 \)[/tex].
Thus, the coefficient of [tex]\( x^2 \)[/tex] in the expansion of [tex]\( (x+2)^3 \)[/tex] is:
[tex]\[
\boxed{6}
\][/tex]
