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We are tasked with finding the coefficient of \( x^2 \) in the expansion of the expression:

\[
(1 + x^2) \left( \frac{x}{2} - \frac{4}{x} \right)^6
\]

### Step 1: Expand \( \left( \frac{x}{2} - \frac{4}{x} \right)^6 \)

We recognize that this is a binomial expansion problem. The binomial theorem states that for \( (a + b)^n \):

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

In this case, let:
- \( a = \frac{x}{2} \)
- \( b = -\frac{4}{x} \)
- \( n = 6 \)

So, we expand \( \left( \frac{x}{2} - \frac{4}{x} \right)^6 \) using the binomial theorem:

\[
\left( \frac{x}{2} - \frac{4}{x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} \left( \frac{x}{2} \right)^{6-k} \left( -\frac{4}{x} \right)^k
\]

Let's simplify a few terms:
\[
\left( \frac{x}{2} \right)^{6-k} = \frac{x^{6-k}}{2^{6-k}}
\]
\[
\left( -\frac{4}{x} \right)^k = (-1)^k \frac{4^k}{x^k}
\]

Thus, each term of the expansion is:

\[
\binom{6}{k} \cdot \frac{x^{6-k}}{2^{6-k}} \cdot (-1)^k \cdot \frac{4^k}{x^k} = \binom{6}{k} \cdot (-1)^k \cdot \frac{4^k}{2^{6-k}} \cdot x^{6-2k}
\]

So the general term of the expansion is:

\[
\binom{6}{k} \cdot (-1)^k \cdot \frac{4^k}{2^{6-k}} \cdot x^{6-2k}
\]

### Step 2: Multiply by \( 1 + x^2 \)

Next, we need to multiply this expansion by \( 1 + x^2 \). For each term of the form \( x^{6-2k} \), multiplying by \( 1 + x^2 \) gives:

\[
x^{6-2k} \cdot (1 + x^2) = x^{6-2k} + x^{8-2k}
\]

### Step 3: Identify the term with \( x^2 \)

We now need to find the values of \( k \) that will result in a term with \( x^2 \).

For the first product, \( x^{6-2k} \), we set \( 6-2k = 2 \), giving:

\[
6 - 2k = 2 \quad \Rightarrow \quad 2k = 4 \quad \Rightarrow \quad k = 2
\]

For the second product, \( x^{8-2k} \), we set \( 8-2k = 2 \), giving:

\[
8 - 2k = 2 \quad \Rightarrow \quad 2k = 6 \quad \Rightarrow \quad k = 3
\]

### Step 4: Calculate the coefficients

We now compute the coefficients for \( k = 2 \) and \( k = 3 \):

- For \( k = 2 \):

\[
\binom{6}{2} \cdot (-1)^2 \cdot \frac{4^2}{2^{6-2}} = \binom{6}{2} \cdot 1 \cdot \frac{16}{2^4} = 15 \cdot \frac{16}{16} = 15
\]

- For \( k = 3 \):

\[
\binom{6}{3} \cdot (-1)^3 \cdot \frac{4^3}{2^{6-3}} = \binom{6}{3} \cdot (-1) \cdot \frac{64}{2^3} = 20 \cdot (-1) \cdot \frac{64}{8} = -20 \cdot 8 = -160
\]

### Step 5: Combine the contributions

The total coefficient of \( x^2 \) is the sum of the contributions from \( k = 2 \) and \( k = 3 \):

\[
\text{Total coefficient} = 15 + (-160) = -145
\]

### Final Answer:
The coefficient of \( x^2 \) in the expansion is \( \boxed{-145} \).

Do you need further details or clarification?

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Here are 5 related questions to expand on this concept:
1. How does the binomial theorem apply to negative exponents?
2. What would be the coefficient of \( x^4 \) in this expansion?
3. How can we use binomial expansions in calculus problems, like integration?
4. What is the general formula for binomial expansion with fractional powers?
5. How would the approach change if the exponent were larger than 6?

**Tip:** When dealing with binomial expansions, always simplify each term before multiplying by additional factors to avoid unnecessary complications.

Find the coefficient of x^2 in the expansion of (1 + x^2) ((x/2) - (4/x))^6.

Explore the detailed process of finding the coefficient of x^2 in the polynomial expansion of (1 + x^2) ((x/2) - (4/x))^6 using the Binomial Theorem. This complex problem involves negative and rational exponents, suitable for advanced high school or early college students interested in algebra.

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