Given that (3x+2) is a factor of P(x) = 18x^3 + bx^2 - x - 2, we set 3x+2=0 to find the root.

Step 1: Apply the Factor Theorem. The Factor Theorem states that if $(ax+b)$ is a factor of a polynomial $P(x)$, then $P\left(-\frac{b}{a}\right) = 0$. Given that $(3x+2)$ is a factor of $P(x) = 18x^3 + bx^2 - x - 2$, we set $3x+2=0$ to find the root. $$3x = -2$$ $$x = -\frac{2}{3}$$ According to the Factor Theorem, $P\left(-\frac{2}{3}\right)$ must be equal to $0$. Step 2: Substitute $x = -\frac{2}{3}$ into $P(x)$ and set the expression to zero. $$P\left(-\frac{2}{3}\right) = 18\left(-\frac{2}{3}\right)^3 + b\left(-\frac{2}{3}\right)^2 - \left(-\frac{2}{3}\right) - 2 = 0$$ Step 3: Simplify the expression. Calculate the powers: $$\left(-\frac{2}{3}\right)^3 = -\frac{2^3}{3^3} = -\frac{8}{27}$$ $$\left(-\frac{2}{3}\right)^2 = \frac{2^2}{3^2} = \frac{4}{9}$$ Substitute these values back into the equation: $$18\left(-\frac{8}{27}\right) + b\left(\frac{4}{9}\right) - \left(-\frac{2}{3}\right) - 2 = 0$$ $$-\frac{18 \cdot 8}{27} + \frac{4b}{9} + \frac{2}{3} - 2 = 0$$ Simplify the first term: $\frac{18}{27} = \frac{2 \cdot 9}{3 \cdot 9} = \frac{2}{3}$. So, $-\frac{2 \cdot 8}{3} = -\frac{16}{3}$. The equation becomes: $$-\frac{16}{3} + \frac{4b}{9} + \frac{2}{3} - 2 = 0$$ Step 4: Solve for $b$. Combine the constant terms: $$\left(-\frac{16}{3} + \frac{2}{3}\right) + \frac{4b}{9} - 2 = 0$$ $$-\frac{14}{3} + \frac{4b}{9} - 2 = 0$$ To eliminate the denominators, multiply the entire equation by the least common multiple of 3 and 9, which is 9: $$9\left(-\frac{14}{3}\right) + 9\left(\frac{4b}{9}\right) - 9(2) = 9(0)$$ $$-3 \cdot 14 + 4b - 18 = 0$$ $$-42 + 4b - 18 = 0$$ Combine the constant terms: $$4b - 60 = 0$$ Add 60 to both sides: $$4b = 60$$ Divide by 4: $$b = \frac{60}{4}$$ $$b = 15$$ The value of $b$ is $\boxed{15}$. 3 done, 2 left today. You're making progress.