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We are asked to find the constant term in the expansion of:

\[
(2 + x)^4 \left(2x + \frac{1}{x}\right)^5
\]

To solve this, we must analyze both binomial expansions separately and identify the combination of terms that will result in a constant (independent of \( x \)) when multiplied together.

### Step 1: Expand \( (2 + x)^4 \)

Using the binomial theorem, we can expand \( (2 + x)^4 \) as follows:

\[
(2 + x)^4 = \sum_{k=0}^{4} \binom{4}{k} 2^{4-k} x^k = 16 + 32x + 24x^2 + 8x^3 + x^4
\]

### Step 2: Expand \( \left(2x + \frac{1}{x}\right)^5 \)

Now, we apply the binomial theorem to expand \( \left(2x + \frac{1}{x}\right)^5 \):

\[
\left(2x + \frac{1}{x}\right)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} \left(\frac{1}{x}\right)^k
\]

This gives the terms:

\[
= 32x^5 + 80x^3 + 80x + 40x^{-1} + 10x^{-3} + x^{-5}
\]

### Step 3: Find combinations to obtain the constant term

Now, to find the constant term, we need to find terms from the expansions of \( (2 + x)^4 \) and \( \left(2x + \frac{1}{x}\right)^5 \) that, when multiplied, give a term where the power of \( x \) is zero.

- From \( 16 \) in \( (2 + x)^4 \) and \( 40x^{-1} \) in \( \left(2x + \frac{1}{x}\right)^5 \), we get:
  \[
  16 \times 40x^{-1} = 640x^{-1}
  \]
- From \( 32x \) in \( (2 + x)^4 \) and \( 80x \) in \( \left(2x + \frac{1}{x}\right)^5 \), we get:
  \[
  32x \times 80x = 2560x^2
  \]
These combinations do not provide

Find the constant term in the expansion of (2 + x)^4 * (2x + 1/x)^5.

Learn how to find the constant term in the binomial expansion of (2 + x)^4 and (2x + 1/x)^5. This problem involves applying the binomial theorem and identifying specific term combinations. Suitable for advanced high school or early college students.

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