1. **Stating the problem:** Solve the equation $$x^5 - x^4 - x^3 - x + 1 = 0$$ for $x$. 2. **Formula and approach:** This is a polynomial equation of degree 5. There is no general formula for solving quintic equations by radicals, so we try to find rational roots using the Rational Root Theorem or factorization. 3. **Rational Root Theorem:** Possible rational roots are factors of the constant term (1) over factors of the leading coefficient (1), so possible roots are $\pm 1$. 4. **Test $x=1$:** $$1^5 - 1^4 - 1^3 - 1 + 1 = 1 - 1 - 1 - 1 + 1 = -1 \neq 0$$ 5. **Test $x=-1$:** $$(-1)^5 - (-1)^4 - (-1)^3 - (-1) + 1 = -1 - 1 + 1 + 1 + 1 = 1 \neq 0$$ 6. Since neither $x=1$ nor $x=-1$ is a root, try to factor by grouping or synthetic division. 7. **Try grouping:** $$x^5 - x^4 - x^3 - x + 1 = (x^5 - x^4 - x^3) + (-x + 1) = x^3(x^2 - x - 1) - (x - 1)$$ 8. Notice $-(x - 1) = -x + 1$, so rewrite: $$x^3(x^2 - x - 1) - (x - 1)$$ 9. Try to factor further: $$= (x - 1)(-1) + x^3(x^2 - x - 1)$$ No obvious factorization emerges, so try polynomial division by $x - 1$: 10. **Divide by $x - 1$:** Using synthetic division with root candidate 1: Coefficients: 1 (x^5), -1 (x^4), -1 (x^3), 0 (x^2), -1 (x), 1 (constant) Carry down 1, multiply by 1, add: 1 | 1 -1 -1 0 -1 1 | 1 0 -1 -1 -2 ---------------- 1 0 -1 -1 -2 -1 Remainder is -1, so $x-1$ is not a factor. 11. **Try $x+1$:** Synthetic division with -1: -1 | 1 -1 -1 0 -1 1 | -1 2 -1 1 0 ---------------- 1 -2 1 -1 0 1 Remainder is 1, so $x+1$ is not a factor. 12. Since no rational roots, the equation must be solved numerically or approximated. 13. **Numerical approximation:** Using numerical methods (e.g., Newton-Raphson), approximate roots can be found. 14. **Summary:** The polynomial has no rational roots. It can be solved numerically or analyzed graphically for approximate roots. **Final answer:** No rational roots; solve numerically for approximate solutions.