To find the \(x^2\) term in the expansion of \((x - 5)^4\) using the Binomial Theorem, we can apply the theorem which states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In this case, \(a = x\), \(b = -5\), and \(n = 4\). We want to find the term that contains \(x^2\), which means we need to find the term where \(n-k=2\) (i.e., the exponent of \(x\) is 2).
Setting \(n-k = 2\):
\[ 4 - k = 2 \implies k = 2 \]
Now, we can find the desired term when \(k = 2\):
\[ \text{Term} = \binom{4}{2} x^{4-2} (-5)^2 \]
Calculating \( \binom{4}{2} \):
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3}{2 \cdot 1} = 6 \]
Now, compute \( (-5)^2 \):
\[ (-5)^2 = 25 \]
Substituting back, we find the \(x^2\) term:
\[ \text{Term} = 6 \cdot x^2 \cdot 25 = 150x^2 \]
Thus, the \(x^2\) term in the expansion of \((x - 5)^4\) is
\[ \boxed{150x^2} \]
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