To find the constant term in the expression $(3 - \frac{1}{2}x)^2 (1 + 2x)^8$, we need to expand each part and look for the term where the variable $x$ has a power of zero (i.e., $x^0$).

Step 1: Expand $(3 - \frac{1}{2}x)^2$

Using the binomial expansion: $(3 - \frac{1}{2}x)^2 = 3^2 - 2 \cdot 3 \cdot \frac{1}{2}x + \left(\frac{1}{2}x\right)^2 = 9 - 3x + \frac{1}{4}x^2.$

Step 2: Expand $(1 + 2x)^8$

Using the binomial theorem: $(1 + 2x)^8 = \sum_{k=0}^{8} \binom{8}{k} \cdot (1)^{8-k} \cdot (2x)^k = \sum_{k=0}^{8} \binom{8}{k} \cdot 2^k x^k.$ We only need terms where the power of $x$ from both expansions add up to zero.

Step 3: Find the Constant Term

The constant term comes from the product of the constant term in $(3 - \frac{1}{2}x)^2$, which is $9$, and the constant term in $(1 + 2x)^8$, which is $1$.

Thus, the constant term is: $9 \cdot 1 = 9.$

Answer

The constant term is 9.

Would you like a more detailed explanation of any step?

1. How does the binomial expansion work for negative exponents?
2. Can we find the coefficient of a specific $x^n$ term?
3. What if the powers in both binomials were higher?
4. How would we handle fractional exponents in binomial expansions?
5. Why does the constant term come only from the product of constant terms in each expansion?

Tip: When looking for constant terms in products, only consider terms that add up to a power of zero.