The question requires solving for the constants $A$ and $B$ in the equation:

$\frac{16}{(x - 2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2}$

Let's break down the steps:

Step 1: Clear the Denominators

To eliminate the fractions, multiply both sides by the Least Common Denominator (LCD), which is $(x - 2)(x + 2)$:

$16 = A(x + 2) + B(x - 2)$

Step 2: Expand the Right Side

Now expand the right side:

$16 = Ax + 2A + Bx - 2B$

Combine like terms:

$16 = (A + B)x + (2A - 2B)$

Step 3: Set Up a System of Equations

Since this equation must hold for all values of $x$, we can equate the coefficients of $x$ and the constant terms on both sides:

1. The coefficient of $x$: $A + B = 0$
2. The constant term: $2A - 2B = 16$

Step 4: Solve the System of Equations

Let's solve for $A$ and $B$ using these equations:

1. From the first equation, $A + B = 0$, we get $B = -A$.

2. Substitute $B = -A$ into the second equation:

   $2A - 2(-A) = 16$

   Simplify:

   $2A + 2A = 16$

   $4A = 16$

   $A = 4$

3. Substitute $A = 4$ back into $B = -A$:

   $B = -4$

Solution

The values of $A$ and $B$ are:

$A = 4 \quad \text{and} \quad B = -4$

Would you like further clarification on any part of this process?

Here are some related questions that build on these concepts:

1. How would you solve for different constants if the equation included additional terms?
2. What would change if the denominators were different, such as $(x - 3)(x + 3)$?
3. How do you interpret the system of equations derived from polynomial identities?
4. Can you apply this method to equations involving more variables or higher-degree polynomials?
5. How would you verify the solution by substituting values back into the original equation?

Tip: When dealing with equations involving fractions, clearing the denominators by multiplying with the LCD is an efficient first step to simplify the equation.