1. We are asked to factor the cubic polynomial $2x^3 + 9x^2 - 6x - 5$. 2. First, try to find rational roots using the Rational Root Theorem. Possible roots are factors of the constant term $-5$ divided by factors of the leading coefficient $2$: $\pm1, \pm\frac{1}{2}, \pm5, \pm\frac{5}{2}$. 3. Test $x=\frac{1}{2}$: $$2\left(\frac{1}{2}\right)^3 + 9\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) - 5 = 2\cdot\frac{1}{8} + 9\cdot\frac{1}{4} - 3 - 5 = \frac{1}{4} + \frac{9}{4} - 8 = \frac{10}{4} - 8 = 2.5 - 8 = -5.5 \neq 0$$ 4. Test $x=-1$: $$2(-1)^3 + 9(-1)^2 - 6(-1) - 5 = -2 + 9 + 6 - 5 = 8 \neq 0$$ 5. Test $x=1$: $$2(1)^3 + 9(1)^2 - 6(1) - 5 = 2 + 9 - 6 - 5 = 0$$ So, $x=1$ is a root. 6. Use polynomial division or synthetic division to divide $2x^3 + 9x^2 - 6x - 5$ by $(x - 1)$: Synthetic division: Coefficients: 2 | 9 | -6 | -5 Bring down 2. Multiply 2 by 1: 2, add to 9: 11. Multiply 11 by 1: 11, add to -6: 5. Multiply 5 by 1: 5, add to -5: 0 (remainder). So quotient is $2x^2 + 11x + 5$. 7. Factor the quadratic $2x^2 + 11x + 5$: Find two numbers that multiply to $2 \times 5 = 10$ and add to $11$: these are $10$ and $1$. Rewrite: $$2x^2 + 10x + x + 5 = 2x(x + 5) + 1(x + 5) = (2x + 1)(x + 5)$$ 8. Therefore, the full factorization is: $$\boxed{(x - 1)(2x + 1)(x + 5)}$$