1. **Problem:** Simplify the expression $$\left(x^{2} + 3x\right)^{2} - 2\left(x^{2} + 3x\right) - 8$$. 2. **Formula and rules:** Recognize this as a quadratic in terms of the expression $u = x^{2} + 3x$. The expression becomes $$u^{2} - 2u - 8$$. 3. **Intermediate work:** Factor the quadratic in $u$: $$u^{2} - 2u - 8 = (u - 4)(u + 2)$$. 4. **Substitute back:** Replace $u$ with $x^{2} + 3x$: $$(x^{2} + 3x - 4)(x^{2} + 3x + 2)$$. 5. **Factor each quadratic:** - For $x^{2} + 3x - 4$, find factors of $-4$ that sum to $3$: $4$ and $-1$. $$x^{2} + 3x - 4 = (x + 4)(x - 1)$$. - For $x^{2} + 3x + 2$, find factors of $2$ that sum to $3$: $1$ and $2$. $$x^{2} + 3x + 2 = (x + 1)(x + 2)$$. 6. **Final factorization:** $$\boxed{(x + 4)(x - 1)(x + 1)(x + 2)}$$. This is the fully factored form of the original expression.