1. **State the problem:** Simplify or factor the polynomial $2x^3 - 3x^2 - 3x + 2$. 2. **Recall the factoring approach:** For cubic polynomials, try to find rational roots using the Rational Root Theorem, then factor by division. 3. **Possible rational roots:** Factors of constant term 2 over factors of leading coefficient 2 are $\pm1, \pm2, \pm\frac{1}{2}$. 4. **Test $x=1$:** $$2(1)^3 - 3(1)^2 - 3(1) + 2 = 2 - 3 - 3 + 2 = -2 \neq 0$$ 5. **Test $x=2$:** $$2(2)^3 - 3(2)^2 - 3(2) + 2 = 2(8) - 3(4) - 6 + 2 = 16 - 12 - 6 + 2 = 0$$ So, $x=2$ is a root. 6. **Divide polynomial by $(x-2)$:** Using synthetic division: Coefficients: 2 | -3 | -3 | 2 Bring down 2. Multiply 2 by 2 = 4, add to -3 = 1. Multiply 1 by 2 = 2, add to -3 = -1. Multiply -1 by 2 = -2, add to 2 = 0 remainder. So quotient is $2x^2 + x -1$. 7. **Factor quadratic $2x^2 + x -1$:** Find two numbers that multiply to $2 \times (-1) = -2$ and add to $1$. These are 2 and -1. Rewrite: $$2x^2 + 2x - x -1 = 2x(x+1) -1(x+1) = (2x -1)(x+1)$$ 8. **Final factorization:** $$2x^3 - 3x^2 - 3x + 2 = (x - 2)(2x - 1)(x + 1)$$ **Answer:** The polynomial factors as $(x - 2)(2x - 1)(x + 1)$.