Find the zeroes of the polynomial` p(x)=(x-2)^(2)-(x+2)^(2)`.

Options

A. 0
B. 2
C. -2
D. 1

Explanation

To find the zeroes of the polynomial \( p(x) = (x-2)^{2} - (x+2)^{2} \), we will follow these steps:

### Step 1: Set the polynomial equal to zero
We start by setting the polynomial equal to zero:
\[ p(x) = (x-2)^{2} - (x+2)^{2} = 0 \]

### Step 2: Use the difference of squares formula
Recognizing that this expression is a difference of squares, we can use the identity \( a^2 - b^2 = (a-b)(a+b) \). Here, let \( a = (x-2) \) and \( b = (x+2) \):
\[ p(x) = \left((x-2) - (x+2)\right)\left((x-2) + (x+2)\right) \]

### Step 3: Simplify the expressions
Now we simplify both factors:
1. For the first factor:
\[ (x-2) - (x+2) = x - 2 - x - 2 = -4 \]
2. For the second factor:
\[ (x-2) + (x+2) = x - 2 + x + 2 = 2x \]

Putting it all together, we have:
\[ p(x) = (-4)(2x) = -8x \]

### Step 4: Set the simplified polynomial to zero
Now we set the simplified polynomial equal to zero:
\[ -8x = 0 \]

### Step 5: Solve for \( x \)
To solve for \( x \), we divide both sides by -8:
\[ x = 0 \]

### Conclusion
Thus, the zero of the polynomial \( p(x) \) is:
\[ \boxed{0} \]