To solve this problem, let's carefully determine the coefficient of $x^3$ in the expansion of $(2 + ax)^4 (5 - ax)$ and equate it to 432.

Step 1: Expansion Method

The given expression is $(2 + ax)^4 (5 - ax)$. The coefficient of $x^3$ comes from the cross-multiplication of terms between $(2 + ax)^4$ and $(5 - ax)$. We will expand both parts using the binomial theorem.

Part 1: Expansion of $(2 + ax)^4$

Using the binomial theorem: $(2 + ax)^4 = \sum_{r=0}^4 \binom{4}{r} (2^{4-r}) (ax)^r$ This gives the terms: $16 + 32(ax) + 24(ax)^2 + 8(ax)^3 + (ax)^4$

Part 2: Expansion of $(5 - ax)$

The linear term is straightforward: $5 - ax$

Step 2: Finding the $x^3$-Term

To get the coefficient of $x^3$, we consider the following contributions:

1. From $8(ax)^3$ in $(2 + ax)^4$, multiplied by $5$ in $(5 - ax)$.
2. From $24(ax)^2$ in $(2 + ax)^4$, multiplied by $-ax$ in $(5 - ax)$.

Contribution 1:

$8(ax)^3 \cdot 5 = 40a^3x^3$

Contribution 2:

$24(ax)^2 \cdot (-ax) = -24a^3x^3$

Step 3: Total Coefficient of $x^3$

Adding these contributions, the coefficient of $x^3$ is: $40a^3 - 24a^3 = 16a^3$

Step 4: Solve for $a$

The coefficient of $x^3$ is given as 432. Therefore: $16a^3 = 432$

Solve for $a^3$: $a^3 = \frac{432}{16} = 27$

Solve for $a$: $a = \sqrt[3]{27} = 3$

Final Answer:

$a = 3$

Would you like further explanation or assistance?
Here are 5 related questions for practice:

1. How would the process change if the coefficient of $x^4$ was given instead?
2. Can you find the coefficient of $x^2$ in the expansion of $(2 + ax)^4 (5 - ax)$?
3. How does the binomial theorem apply to expansions involving multiple variables?
4. If $a = 3$, what is the constant term in the expansion of $(2 + ax)^4 (5 - ax)$?
5. How do higher-order terms affect the overall expansion of a polynomial?

Tip: When dealing with binomial expansions, always identify which terms contribute to the specific power you are solving for to streamline calculations.