Factor Polynomial 7B4Be0

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 $2$. So, $$y^2 - 2y - 8 = (y - 4)(y + 2)$$ 5. **Substitute back $y = x^2$:** $$(x^2 - 4)(x^2 + 2)$$ 6. **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. 7. **Final factorization:** $$x^4 - 2x^2 - 8 = (x - 2)(x + 2)(x^2 + 2)$$