1. **State the problem:** Find the point(s) where the tangent to the curve $$y = (x^2 + 2x + 1)(x^2 + 2x + 1)$$ is horizontal. 2. **Rewrite the function:** Notice that $$y = (x^2 + 2x + 1)^2 = (x+1)^4$$. 3. **Find the derivative:** The slope of the tangent line is given by $$y' = \frac{dy}{dx}$$. Using the chain rule: $$y' = 4(x+1)^3$$ 4. **Set the derivative equal to zero to find horizontal tangents:** $$4(x+1)^3 = 0$$ Divide both sides by 4: $$\cancel{4}(x+1)^3 = 0 \Rightarrow (x+1)^3 = 0$$ 5. **Solve for x:** $$x + 1 = 0 \Rightarrow x = -1$$ 6. **Find the corresponding y-coordinate:** $$y = (x+1)^4 = (-1+1)^4 = 0^4 = 0$$ 7. **Conclusion:** The tangent is horizontal at the point $$(-1, 0)$$. **Final answer:** $$(-1, 0)$$