1. The problem is to factor the cubic polynomial $x^3 + x + 1$. 2. To factor a cubic polynomial, we first check if it has any rational roots using the Rational Root Theorem. Possible roots are factors of the constant term (1) over factors of the leading coefficient (1), so possible roots are $\pm 1$. 3. Test $x=1$: $1^3 + 1 + 1 = 3 \neq 0$. 4. Test $x=-1$: $(-1)^3 + (-1) + 1 = -1 -1 + 1 = -1 \neq 0$. 5. Since there are no rational roots, the polynomial cannot be factored over the rationals into linear factors. 6. We can try to factor it as a product of a linear and a quadratic polynomial: $$(x - a)(x^2 + bx + c) = x^3 + x + 1$$ 7. Expanding the right side: $$x^3 + bx^2 + cx - a x^2 - a b x - a c = x^3 + (b - a) x^2 + (c - a b) x - a c$$ 8. Equate coefficients with $x^3 + 0 x^2 + 1 x + 1$: - Coefficient of $x^2$: $b - a = 0 \Rightarrow b = a$ - Coefficient of $x$: $c - a b = 1 \Rightarrow c - a^2 = 1$ - Constant term: $-a c = 1 \Rightarrow c = -\frac{1}{a}$ 9. Substitute $c$ into the $x$ coefficient equation: $$-\frac{1}{a} - a^2 = 1 \Rightarrow -\frac{1}{a} = 1 + a^2 \Rightarrow -1 = a (1 + a^2) \Rightarrow a^3 + a + 1 = 0$$ 10. The value $a$ satisfies the original cubic equation, so the factorization is: $$x^3 + x + 1 = (x - a)(x^2 + a x - \frac{1}{a})$$ 11. Since $a$ is a root of the cubic, the factorization is in terms of $a$ which is irrational or complex. 12. Therefore, the polynomial is irreducible over the rationals and factors only with roots of the cubic. Final answer: The polynomial $x^3 + x + 1$ has no rational factors and factors as $$(x - a)(x^2 + a x - \frac{1}{a})$$ where $a$ is a root of $x^3 + x + 1 = 0$.