1. State the problem: Find the x-intercepts of the function $$f(x)=3x^3+2x^2-5x-2$$. The x-intercepts occur where $$f(x)=0$$. 2. Set the equation to zero: $$3x^3+2x^2-5x-2=0$$. 3. Attempt to factor or use rational root theorem to find roots. Test possible rational roots $$\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$$. 4. Test $$x=1$$: $$3(1)^3 + 2(1)^2 - 5(1) - 2 = 3 + 2 - 5 - 2 = -2 \neq 0$$. 5. Test $$x=-1$$: $$3(-1)^3 + 2(-1)^2 - 5(-1) - 2 = -3 + 2 + 5 - 2 = 2 \neq 0$$. 6. Test $$x=2$$: $$3(2)^3 + 2(2)^2 - 5(2) - 2 = 3(8) + 2(4) - 10 - 2 = 24 + 8 - 12 = 20 \neq 0$$. 7. Test $$x=-2$$: $$3(-2)^3 + 2(-2)^2 - 5(-2) - 2 = 3(-8) + 2(4) + 10 - 2 = -24 + 8 + 10 - 2 = -8 \neq 0$$. 8. Test $$x=\frac{1}{3}$$: $$3\left(\frac{1}{3}\right)^3 + 2\left(\frac{1}{3}\right)^2 - 5\left(\frac{1}{3}\right) - 2 = 3\left(\frac{1}{27}\right) + 2\left(\frac{1}{9}\right) - \frac{5}{3} - 2 = \frac{1}{9} + \frac{2}{9} - \frac{5}{3} - 2 = \frac{3}{9} - \frac{5}{3} - 2 = \frac{1}{3} - \frac{5}{3} - 2 = -\frac{4}{3} - 2 \neq 0$$. 9. Test $$x=-\frac{1}{3}$$: $$3\left(-\frac{1}{3}\right)^3 + 2\left(-\frac{1}{3}\right)^2 - 5\left(-\frac{1}{3}\right) - 2 = 3\left(-\frac{1}{27}\right) + 2\left(\frac{1}{9}\right) + \frac{5}{3} - 2 = -\frac{1}{9} + \frac{2}{9} + \frac{5}{3} - 2 = \frac{1}{9} + \frac{5}{3} - 2 = \frac{1}{9} + \frac{15}{9} - \frac{18}{9} = -\frac{2}{9} \neq 0$$. 10. Since none of the rational candidates work, use the cubic formula or numerical methods to find roots. 11. Use approximate methods (such as Newton-Raphson or graphing) to estimate roots. 12. By inspection or graphing, approximate roots are around $$x \approx 1.1$$, $$x \approx -1.5$$ and $$x \approx -0.45$$. 13. Confirm via numerical solving software or calculator. 14. Final answer: The x-intercepts (roots) of $$f(x) = 3x^3 + 2x^2 -5x - 2$$ are approximately $$x \approx 1.1, -1.5, -0.45$$.