1. The problem is to solve the algebraic expression $3y^3 - 2y^2 + 5y - 6$ by factoring or finding its roots. 2. We use the factoring method to solve cubic expressions. The goal is to find values of $y$ that make the expression equal to zero. 3. First, try to find rational roots using the Rational Root Theorem. Possible roots are factors of the constant term 6: $\pm1, \pm2, \pm3, \pm6$. 4. Test $y=1$: $3(1)^3 - 2(1)^2 + 5(1) - 6 = 3 - 2 + 5 - 6 = 0$. So, $y=1$ is a root. 5. Divide the polynomial by $(y-1)$ using synthetic division or long division: $$3y^3 - 2y^2 + 5y - 6 \div (y-1) = 3y^2 + y + 6$$ 6. Now solve the quadratic $3y^2 + y + 6 = 0$ using the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \times 3 \times 6}}{2 \times 3} = \frac{-1 \pm \sqrt{1 - 72}}{6} = \frac{-1 \pm \sqrt{-71}}{6}$$ 7. Since the discriminant is negative ($-71$), the quadratic has no real roots, only complex roots: $$y = \frac{-1 \pm i\sqrt{71}}{6}$$ 8. Final answer: The expression equals zero at $y=1$ (real root) and at $y=\frac{-1 \pm i\sqrt{71}}{6}$ (complex roots).