1. **State the problem:** Simplify or analyze the polynomial expression $3x^3 - x^2 + 6x - 2$. 2. **Identify the polynomial:** This is a cubic polynomial of the form $ax^3 + bx^2 + cx + d$ where $a=3$, $b=-1$, $c=6$, and $d=-2$. 3. **Check for possible factorization:** We can try to factor the polynomial by looking for rational roots using the Rational Root Theorem. Possible roots are factors of $d$ over factors of $a$: $\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$. 4. **Test $x=1$:** $$3(1)^3 - (1)^2 + 6(1) - 2 = 3 - 1 + 6 - 2 = 6 \neq 0$$ 5. **Test $x=-1$:** $$3(-1)^3 - (-1)^2 + 6(-1) - 2 = -3 - 1 - 6 - 2 = -12 \neq 0$$ 6. **Test $x=2$:** $$3(2)^3 - (2)^2 + 6(2) - 2 = 3(8) - 4 + 12 - 2 = 24 - 4 + 12 - 2 = 30 \neq 0$$ 7. **Test $x=-2$:** $$3(-2)^3 - (-2)^2 + 6(-2) - 2 = 3(-8) - 4 - 12 - 2 = -24 - 4 - 12 - 2 = -42 \neq 0$$ 8. **Test $x=\frac{1}{3}$:** $$3\left(\frac{1}{3}\right)^3 - \left(\frac{1}{3}\right)^2 + 6\left(\frac{1}{3}\right) - 2 = 3\left(\frac{1}{27}\right) - \frac{1}{9} + 2 - 2 = \frac{1}{9} - \frac{1}{9} + 0 = 0$$ 9. Since $x=\frac{1}{3}$ is a root, factor out $(x - \frac{1}{3})$: $$3x^3 - x^2 + 6x - 2 = (x - \frac{1}{3})(\text{quadratic})$$ Multiply both sides by 3 to clear fraction: $$3x^3 - x^2 + 6x - 2 = \left(3x - 1\right)(\text{quadratic})$$ 10. Use polynomial division or synthetic division to find the quadratic factor: Divide $3x^3 - x^2 + 6x - 2$ by $3x - 1$: $$\frac{3x^3 - x^2 + 6x - 2}{3x - 1} = x^2 + 0x + 2$$ 11. So the factorization is: $$3x^3 - x^2 + 6x - 2 = (3x - 1)(x^2 + 2)$$ 12. The quadratic $x^2 + 2$ has no real roots since $x^2 = -2$ has no real solution. **Final answer:** $$3x^3 - x^2 + 6x - 2 = (3x - 1)(x^2 + 2)$$