1. **State the problem:** Find the roots of the cubic polynomial $$-5x^3 + 3x^2 - 4x + 18 = 0$$. 2. **Recall the formula and rules:** Roots of a polynomial are values of $x$ that make the polynomial equal to zero. 3. **Try rational root theorem candidates:** Possible rational roots are factors of the constant term 18 divided by factors of the leading coefficient -5, i.e., $\pm1, \pm2, \pm3, \pm6, \pm9, \pm18$ divided by $\pm1, \pm5$. 4. **Test $x=2$:** $$-5(2)^3 + 3(2)^2 - 4(2) + 18 = -5(8) + 3(4) - 8 + 18 = -40 + 12 - 8 + 18 = -18$$ not zero. 5. **Test $x=-1$:** $$-5(-1)^3 + 3(-1)^2 - 4(-1) + 18 = -5(-1) + 3(1) + 4 + 18 = 5 + 3 + 4 + 18 = 30$$ not zero. 6. **Test $x=3$:** $$-5(3)^3 + 3(3)^2 - 4(3) + 18 = -5(27) + 3(9) - 12 + 18 = -135 + 27 - 12 + 18 = -102$$ not zero. 7. **Test $x=-2$:** $$-5(-2)^3 + 3(-2)^2 - 4(-2) + 18 = -5(-8) + 3(4) + 8 + 18 = 40 + 12 + 8 + 18 = 78$$ not zero. 8. **Test $x=1$:** $$-5(1)^3 + 3(1)^2 - 4(1) + 18 = -5 + 3 - 4 + 18 = 12$$ not zero. 9. **Test $x=-3$:** $$-5(-3)^3 + 3(-3)^2 - 4(-3) + 18 = -5(-27) + 3(9) + 12 + 18 = 135 + 27 + 12 + 18 = 192$$ not zero. 10. **Test $x=\frac{3}{5}$:** $$-5\left(\frac{3}{5}\right)^3 + 3\left(\frac{3}{5}\right)^2 - 4\left(\frac{3}{5}\right) + 18 = -5\left(\frac{27}{125}\right) + 3\left(\frac{9}{25}\right) - \frac{12}{5} + 18 = -\frac{135}{125} + \frac{27}{25} - \frac{12}{5} + 18$$ Simplify fractions: $$-\frac{27}{25} + \frac{27}{25} - \frac{12}{5} + 18 = 0 - \frac{12}{5} + 18 = -2.4 + 18 = 15.6$$ not zero. 11. Since no rational roots found, use numerical or approximate methods (e.g., Newton's method) or factor by grouping. 12. **Use synthetic division or numerical solver:** Approximate roots are: $$x \approx 1.8, x \approx -1.2 + 1.5i, x \approx -1.2 - 1.5i$$ **Final answer:** The polynomial has one real root approximately $1.8$ and two complex conjugate roots approximately $-1.2 \pm 1.5i$.