To determine where the graph of $P(x) = x^4 - 16$ crosses the $x$-axis, we need to find the real roots of the polynomial $P(x) = 0$.

Step 1: Solve $P(x) = 0$

$P(x) = x^4 - 16 = 0$

Rewriting this as a difference of squares: $x^4 - 16 = (x^2 - 4)(x^2 + 4)$

Step 2: Factor further

The term $x^2 - 4$ can be factored again: $x^2 - 4 = (x - 2)(x + 2)$

So: $P(x) = (x - 2)(x + 2)(x^2 + 4)$

Step 3: Solve for the roots

From the factorization, the roots of $P(x)$ are: $x - 2 = 0 \quad \implies \quad x = 2$ $x + 2 = 0 \quad \implies \quad x = -2$ $x^2 + 4 = 0 \quad \implies \quad x^2 = -4 \quad \implies \quad x = \pm 2i$

Thus, the roots are $x = 2, -2, 2i, -2i$.

Step 4: Identify $x$-intercepts

The graph of $P(x)$ crosses the $x$-axis only at real roots. Hence, the graph crosses the $x$-axis at: $x = 2 \quad \text{and} \quad x = -2$

Final Answer:

The correct option is: $\boxed{2, -2}$

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Would you like further explanation on factoring or interpreting the roots? Here are related questions for practice:

1. How do you factor higher-degree polynomials like $x^4 - 16$?
2. How do you solve quadratic equations with complex roots?
3. Why do complex roots not correspond to $x$-intercepts?
4. What is the difference between a polynomial's roots and its $x$-intercepts?
5. Can you describe the shape of the graph of $P(x) = x^4 - 16$?

Tip: Remember, a polynomial of degree $n$ has exactly $n$ roots, including real and complex ones!