Find all zeros (real and complex). Factor the polynomial as a product of linear factors. P(x)=x^4 -225

We are given the polynomial \( P(x) = x^4 - 225 \), and we need to find its zeros and factor it as a product of linear factors.

### Step 1: Recognize the polynomial as a difference of squares
We can rewrite the polynomial as a difference of squares:

\[
P(x) = x^4 - 225 = (x^2)^2 - 15^2
\]

This can be factored as:

\[
P(x) = (x^2 - 15)(x^2 + 15)
\]

### Step 2: Find the zeros of each factor

#### For \( x^2 - 15 = 0 \):
\[
x^2 = 15
\]
\[
x = \pm \sqrt{15}
\]

So, two real zeros are \( x = \sqrt{15} \) and \( x = -\sqrt{15} \).

#### For \( x^2 + 15 = 0 \):
\[
x^2 = -15
\]
\[
x = \pm \sqrt{-15} = \pm \sqrt{15}i
\]

So, two complex zeros are \( x = \sqrt{15}i \) and \( x = -\sqrt{15}i \).

### Step 3: Factor the polynomial as a product of linear factors
The polynomial can now be factored fully as:

\[
P(x) = (x - \sqrt{15})(x + \sqrt{15})(x - \sqrt{15}i)(x + \sqrt{15}i)
\]

### Final Answer:
- **Zeros**: \( x = \sqrt{15}, x = -\sqrt{15}, x = \sqrt{15}i, x = -\sqrt{15}i \)
- **Factorization**:
\[
P(x) = (x - \sqrt{15})(x + \sqrt{15})(x - \sqrt{15}i)(x + \sqrt{15}i)
\]

Would you like further details or have any questions? Here are some related questions:

1. How can the difference of squares method be applied to other types of polynomials?
2. Can you explain how to find complex roots for higher-degree polynomials?
3. What are the properties of complex numbers and their conjugates?
4. How does factoring help in solving polynomial equations?
5. What is the relationship between the degree of a polynomial and the number of its zeros?

**Tip**: Always check for the possibility of using the difference of squares when you see an expression involving two terms, especially if both are perfect squares.

Learn how to factor the polynomial P(x) = x^4 - 225 and find all real and complex zeros. This detailed solution uses the difference of squares method and introduces concepts like complex roots. Suitable for Grades 10-12.

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