Question: Factor completely over the integers. $x^4 - 2x^2 - 8$
1. **State the problem:** Factor the polynomial $$x^4 - 2x^2 - 8$$ completely over the integers.
2. **Identify the form:** Notice that the polynomial is in terms of $x^4$, $x^2$, and a constant. We can treat it as a quadratic in $x^2$ by letting $y = x^2$.
3. **Rewrite the polynomial:**
$$x^4 - 2x^2 - 8 = y^2 - 2y - 8$$
4. **Factor the quadratic in $y$:**
We look for two numbers that multiply to $-8$ and add to $-2$. These numbers are $-4$ and $2$.
5. **Factorization:**
$$y^2 - 2y - 8 = (y - 4)(y + 2)$$
6. **Substitute back $y = x^2$:**
$$(x^2 - 4)(x^2 + 2)$$
7. **Factor further if possible:**
$x^2 - 4$ is a difference of squares:
$$x^2 - 4 = (x - 2)(x + 2)$$
$x^2 + 2$ cannot be factored further over the integers.
8. **Final factorization:**
$$(x - 2)(x + 2)(x^2 + 2)$$