To determine the constant term in the expansion of the binomial [tex]\((x-2)^4\)[/tex], we can use the Binomial Theorem. The Binomial Theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\][/tex]
In this case, [tex]\(a = x\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(n = 4\)[/tex]. The expansion of [tex]\((x-2)^4\)[/tex] can be written as:
[tex]\[
(x-2)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} (-2)^k
\][/tex]
We are specifically interested in the constant term in the expansion. The constant term is the term where there is no [tex]\(x\)[/tex] present, meaning [tex]\(x\)[/tex] must be raised to the power of 0. For this to happen, we need [tex]\(4-k = 0\)[/tex], that is, [tex]\(k = 4\)[/tex].
With [tex]\(k = 4\)[/tex], the constant term is:
[tex]\[
\binom{4}{4} x^{4-4} (-2)^4
\][/tex]
Simplifying each part, we have:
[tex]\[
\binom{4}{4} = 1
\][/tex]
[tex]\[
x^0 = 1
\][/tex]
[tex]\[
(-2)^4 = 16
\][/tex]
Therefore, the constant term is:
[tex]\[
1 \cdot 1 \cdot 16 = 16
\][/tex]
So, the constant term in the expansion of [tex]\((x-2)^4\)[/tex] is:
[tex]\[
\boxed{16}
\][/tex]
Hence, the correct answer is B. 16.
