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To solve this problem, let's go through the steps required to find the zeros and factor the polynomial \( P(x) = x^4 - 81 \).

### Step 1: Recognize the Polynomial as a Difference of Squares
We can rewrite \( x^4 - 81 \) as:
\[
P(x) = (x^2)^2 - 9^2
\]
This is a difference of squares, which can be factored as:
\[
P(x) = (x^2 - 9)(x^2 + 9)
\]

### Step 2: Factor Each Quadratic Term Further
1. **Factor \( x^2 - 9 \):**
   \[
   x^2 - 9 = (x - 3)(x + 3)
   \]
2. **Factor \( x^2 + 9 \):**
   Since \( x^2 + 9 \) has complex roots, we use:
   \[
   x^2 + 9 = (x - 3i)(x + 3i)
   \]

### Step 3: Write \( P(x) \) as a Product of Linear Factors
Combining these factors, we get:
\[
P(x) = (x - 3)(x + 3)(x - 3i)(x + 3i)
\]

### Step 4: List All Zeros
From the factorization, the zeros (solutions) of \( P(x) = 0 \) are:
\[
x_1 = -3, \quad x_2 = 3, \quad x_3 = -3i, \quad x_4 = 3i
\]

### Final Answer
The polynomial \( P(x) \) factored as a product of linear factors is:
\[
P(x) = (x - 3)(x + 3)(x - 3i)(x + 3i)
\]

Would you like more details or have any questions on this?

### Related Questions
1. How do you factor other polynomials that are differences of squares?
2. What are the steps to factor polynomials with complex roots?
3. Why does \( x^2 + 9 \) not factor further over the real numbers?
4. How can you find zeros of polynomials with higher degrees (greater than 4)?
5. How do you use the Fundamental Theorem of Algebra to determine the number of roots?

### Tip
When factoring polynomials, always check for special patterns like difference of squares, as they simplify the process greatly.

Find all zeros (real and complex) of the polynomial P(x) = x^4 - 81, and factor it as a product of linear factors.

Learn how to factor the polynomial P(x) = x^4 - 81 and find all zeros, both real and complex. This problem involves polynomial factorization using the difference of squares and complex number concepts, and is suitable for students in Grades 10-12.

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