1. **State the problem:** Factorize the polynomial $x^4 - 13x^2 - 36$. 2. **Identify the type of polynomial:** This is a quartic polynomial in terms of $x$, but notice it can be treated as a quadratic in $x^2$ because the powers are $4$, $2$, and $0$. 3. **Rewrite the polynomial:** Let $y = x^2$. Then the polynomial becomes $y^2 - 13y - 36$. 4. **Factor the quadratic in $y$:** We look for two numbers that multiply to $-36$ and add to $-13$. These numbers are $-9$ and $4$. 5. **Write the factorization:** $$y^2 - 13y - 36 = (y - 9)(y + 4)$$ 6. **Substitute back $y = x^2$:** $$(x^2 - 9)(x^2 + 4)$$ 7. **Factor further if possible:** $x^2 - 9$ is a difference of squares: $$(x - 3)(x + 3)$$ $x^2 + 4$ cannot be factored further over the real numbers. 8. **Final factorization:** $$(x - 3)(x + 3)(x^2 + 4)$$