1. **State the problem:** Solve the equation $$2x + \frac{(x-1)^2 (128x) - (64x^2 + 2x - 2)}{(x-1)^4} = 0$$ for $x$. 2. **Rewrite the equation:** To combine terms, write the first term $2x$ with denominator $(x-1)^4$: $$2x = \frac{2x (x-1)^4}{(x-1)^4}$$ So the equation becomes: $$\frac{2x (x-1)^4 + (x-1)^2 (128x) - (64x^2 + 2x - 2)}{(x-1)^4} = 0$$ 3. **Set numerator equal to zero:** Since denominator $(x-1)^4 \neq 0$ for $x \neq 1$, solve: $$2x (x-1)^4 + (x-1)^2 (128x) - (64x^2 + 2x - 2) = 0$$ 4. **Factor powers of $(x-1)^2$ where possible:** $$2x (x-1)^4 + 128x (x-1)^2 - (64x^2 + 2x - 2) = 0$$ Rewrite $2x (x-1)^4$ as $2x (x-1)^2 (x-1)^2$: $$2x (x-1)^2 (x-1)^2 + 128x (x-1)^2 - (64x^2 + 2x - 2) = 0$$ Group terms with $(x-1)^2$: $$(x-1)^2 (2x (x-1)^2 + 128x) - (64x^2 + 2x - 2) = 0$$ 5. **Expand $(x-1)^2 = x^2 - 2x + 1$ and $(x-1)^4 = (x-1)^2 \cdot (x-1)^2$ if needed:** Calculate $2x (x-1)^2$: $$2x (x^2 - 2x + 1) = 2x^3 - 4x^2 + 2x$$ So inside the parentheses: $$2x (x-1)^2 + 128x = (2x^3 - 4x^2 + 2x) + 128x = 2x^3 - 4x^2 + 130x$$ 6. **Rewrite the equation:** $$(x-1)^2 (2x^3 - 4x^2 + 130x) - (64x^2 + 2x - 2) = 0$$ 7. **Expand $(x-1)^2 (2x^3 - 4x^2 + 130x)$:** $$(x^2 - 2x + 1)(2x^3 - 4x^2 + 130x)$$ Multiply term by term: - $x^2 \cdot 2x^3 = 2x^5$ - $x^2 \cdot (-4x^2) = -4x^4$ - $x^2 \cdot 130x = 130x^3$ - $-2x \cdot 2x^3 = -4x^4$ - $-2x \cdot (-4x^2) = 8x^3$ - $-2x \cdot 130x = -260x^2$ - $1 \cdot 2x^3 = 2x^3$ - $1 \cdot (-4x^2) = -4x^2$ - $1 \cdot 130x = 130x$ Sum all: $$2x^5 - 4x^4 + 130x^3 - 4x^4 + 8x^3 - 260x^2 + 2x^3 - 4x^2 + 130x$$ Combine like terms: - $x^5$: $2x^5$ - $x^4$: $-4x^4 - 4x^4 = -8x^4$ - $x^3$: $130x^3 + 8x^3 + 2x^3 = 140x^3$ - $x^2$: $-260x^2 - 4x^2 = -264x^2$ - $x$: $130x$ So: $$2x^5 - 8x^4 + 140x^3 - 264x^2 + 130x$$ 8. **Substitute back into the equation:** $$2x^5 - 8x^4 + 140x^3 - 264x^2 + 130x - (64x^2 + 2x - 2) = 0$$ Distribute minus: $$2x^5 - 8x^4 + 140x^3 - 264x^2 + 130x - 64x^2 - 2x + 2 = 0$$ Combine like terms: - $x^2$: $-264x^2 - 64x^2 = -328x^2$ - $x$: $130x - 2x = 128x$ Final polynomial: $$2x^5 - 8x^4 + 140x^3 - 328x^2 + 128x + 2 = 0$$ 9. **Solve the polynomial:** Try rational roots using factors of constant term 2 and leading coefficient 2: possible roots $\pm1, \pm2, \pm \frac{1}{2}$. Test $x=1$: $$2(1)^5 - 8(1)^4 + 140(1)^3 - 328(1)^2 + 128(1) + 2 = 2 - 8 + 140 - 328 + 128 + 2 = -64 \neq 0$$ Test $x=2$: $$2(32) - 8(16) + 140(8) - 328(4) + 128(2) + 2 = 64 - 128 + 1120 - 1312 + 256 + 2 = 2 \neq 0$$ Test $x=\frac{1}{2}$: Calculate each term: - $2(\frac{1}{2})^5 = 2 \cdot \frac{1}{32} = \frac{1}{16}$ - $-8(\frac{1}{2})^4 = -8 \cdot \frac{1}{16} = -\frac{1}{2}$ - $140(\frac{1}{2})^3 = 140 \cdot \frac{1}{8} = 17.5$ - $-328(\frac{1}{2})^2 = -328 \cdot \frac{1}{4} = -82$ - $128(\frac{1}{2}) = 64$ - $+2$ Sum: $$\frac{1}{16} - \frac{1}{2} + 17.5 - 82 + 64 + 2 = 0.0625 - 0.5 + 17.5 - 82 + 64 + 2 = 1.0625 \neq 0$$ Test $x=-1$: $$2(-1)^5 - 8(-1)^4 + 140(-1)^3 - 328(-1)^2 + 128(-1) + 2 = -2 - 8 - 140 - 328 - 128 + 2 = -604 \neq 0$$ Test $x=-\frac{1}{2}$: Calculate each term: - $2(-\frac{1}{2})^5 = 2 \cdot (-\frac{1}{32}) = -\frac{1}{16}$ - $-8(-\frac{1}{2})^4 = -8 \cdot \frac{1}{16} = -\frac{1}{2}$ - $140(-\frac{1}{2})^3 = 140 \cdot (-\frac{1}{8}) = -17.5$ - $-328(-\frac{1}{2})^2 = -328 \cdot \frac{1}{4} = -82$ - $128(-\frac{1}{2}) = -64$ - $+2$ Sum: $$-\frac{1}{16} - \frac{1}{2} - 17.5 - 82 - 64 + 2 = -162.0625 \neq 0$$ 10. **No simple rational roots found; use numerical or graphical methods to approximate roots.** **Final answer:** The solutions to the original equation satisfy the polynomial: $$2x^5 - 8x^4 + 140x^3 - 328x^2 + 128x + 2 = 0$$ which can be solved numerically.