1. **State the problem:** Solve the equation $$2x + \frac{138x^3 - 192x^2 + 126x + 2}{(x-1)^4} = 0.$$\n\n2. **Rewrite the equation:** Multiply both sides by $(x-1)^4$ to clear the denominator:\n$$2x(x-1)^4 + 138x^3 - 192x^2 + 126x + 2 = 0.$$\n\n3. **Expand $(x-1)^4$:** Using the binomial expansion,\n$$(x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1.$$\n\n4. **Multiply $2x$ by the expansion:**\n$$2x(x^4 - 4x^3 + 6x^2 - 4x + 1) = 2x^5 - 8x^4 + 12x^3 - 8x^2 + 2x.$$\n\n5. **Substitute back into the equation:**\n$$2x^5 - 8x^4 + 12x^3 - 8x^2 + 2x + 138x^3 - 192x^2 + 126x + 2 = 0.$$\n\n6. **Combine like terms:**\n$$2x^5 - 8x^4 + (12x^3 + 138x^3) + (-8x^2 - 192x^2) + (2x + 126x) + 2 = 0,$$\nwhich simplifies to\n$$2x^5 - 8x^4 + 150x^3 - 200x^2 + 128x + 2 = 0.$$\n\n7. **Divide entire equation by 2 to simplify:**\n$$\cancel{2}x^5 - \cancel{2}4x^4 + 75x^3 - 100x^2 + 64x + 1 = 0.$$\n\n8. **Final simplified polynomial equation:**\n$$x^5 - 4x^4 + 75x^3 - 100x^2 + 64x + 1 = 0.$$\n\n9. **Solve the polynomial:** This is a quintic equation and may require numerical methods or graphing to find roots.\n\n**Answer:** The solutions are the roots of $$x^5 - 4x^4 + 75x^3 - 100x^2 + 64x + 1 = 0.$$