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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? 

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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.

To solve for A and B, multiply each side of the equation 16 / ((x - 2)(x + 2)) = A / (x - 2) + B / (x + 2) by the LCD and simplify.

Learn how to solve for constants A and B in the equation 16 / ((x - 2)(x + 2)) = A / (x - 2) + B / (x + 2) by clearing denominators and setting up a system of equations. This algebraic solution includes the concepts of partial fraction decomposition and polynomial identities. Suitable for high school students in Grades 10-12.

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