Question: Factor completely over the integers. $x^4 - 7x^2 - 18$
1. **State the problem:** Factor the polynomial $$x^4 - 7x^2 - 18$$ completely over the integers.
2. **Recognize the structure:** Notice that the polynomial is in terms of $x^4$ and $x^2$. Let us use substitution to simplify.
3. **Substitution:** Let $$y = x^2$$. Then the polynomial becomes:
$$y^2 - 7y - 18$$
4. **Factor the quadratic in $y$:** We look for two numbers that multiply to $$-18$$ and add to $$-7$$. These numbers are $$-9$$ and $$2$$.
5. **Write the factorization:**
$$y^2 - 7y - 18 = (y - 9)(y + 2)$$
6. **Back-substitute $y = x^2$:**
$$(x^2 - 9)(x^2 + 2)$$
7. **Factor further if possible:**
$$x^2 - 9$$ is a difference of squares:
$$x^2 - 9 = (x - 3)(x + 3)$$
8. **Final factorization:**
$$(x - 3)(x + 3)(x^2 + 2)$$
9. **Check for irreducibility:**
The quadratic $$x^2 + 2$$ cannot be factored further over the integers.
**Answer:**
$$(x - 3)(x + 3)(x^2 + 2)$$