1. **State the problem:** Simplify or analyze the cubic polynomial $2x^3 - 3x^2 - 11x + 6$. 2. **Formula and rules:** To factor a cubic polynomial, we can try to find rational roots using the Rational Root Theorem, which states possible roots are factors of the constant term divided by factors of the leading coefficient. 3. **Apply Rational Root Theorem:** Possible roots are $\pm1, \pm2, \pm3, \pm6$ divided by $\pm1, \pm2$, so possible roots are $\pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm\frac{3}{2}, \pm6$. 4. **Test roots:** Evaluate polynomial at these values: - At $x=1$: $2(1)^3 - 3(1)^2 - 11(1) + 6 = 2 - 3 - 11 + 6 = -6$ (not zero) - At $x=2$: $2(8) - 3(4) - 11(2) + 6 = 16 - 12 - 22 + 6 = -12$ (not zero) - At $x=3$: $2(27) - 3(9) - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ (root found) 5. **Factor out $(x-3)$:** Use polynomial division or synthetic division to divide $2x^3 - 3x^2 - 11x + 6$ by $(x-3)$: $$\frac{2x^3 - 3x^2 - 11x + 6}{x - 3} = 2x^2 + 3x - 2$$ 6. **Factor quadratic $2x^2 + 3x - 2$:** Find two numbers that multiply to $2 \times (-2) = -4$ and add to $3$; these are $4$ and $-1$. Rewrite: $$2x^2 + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2)$$ 7. **Final factorization:** $$2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x - 1)(x + 2)$$ **Answer:** The polynomial factors as $(x - 3)(2x - 1)(x + 2)$.