1. **State the problem:** Factorize the cubic polynomial equation $$2X^3 - 3X + 1 = 0$$. 2. **Recall the factorization approach:** For cubic polynomials, try to find rational roots using the Rational Root Theorem, then factor by division. 3. **Apply the Rational Root Theorem:** Possible rational roots are factors of the constant term (1) over factors of the leading coefficient (2), i.e., $$\pm 1, \pm \frac{1}{2}$$. 4. **Test roots:** - For $$X=1$$: $$2(1)^3 - 3(1) + 1 = 2 - 3 + 1 = 0$$, so $$X=1$$ is a root. 5. **Divide the polynomial by $$X-1$$:** Use polynomial division or synthetic division: $$2X^3 - 3X + 1 \div (X - 1) = 2X^2 + 2X - 1$$. 6. **Factor the quadratic:** $$2X^2 + 2X - 1 = 0$$ Use the quadratic formula: $$X = \frac{-2 \pm \sqrt{(2)^2 - 4(2)(-1)}}{2 \times 2} = \frac{-2 \pm \sqrt{4 + 8}}{4} = \frac{-2 \pm \sqrt{12}}{4} = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-1 \pm \sqrt{3}}{2}$$. 7. **Write the full factorization:** $$2X^3 - 3X + 1 = (X - 1)\left(2X^2 + 2X - 1\right) = (X - 1)\left(2X - (-1 + \sqrt{3})\right)\left(2X - (-1 - \sqrt{3})\right)$$ or equivalently: $$ (X - 1)\left(X - \frac{-1 + \sqrt{3}}{2}\right)\left(X - \frac{-1 - \sqrt{3}}{2}\right) $$. **Final answer:** $$2X^3 - 3X + 1 = (X - 1)\left(X - \frac{-1 + \sqrt{3}}{2}\right)\left(X - \frac{-1 - \sqrt{3}}{2}\right)$$.