Find the value of k, if 2x + 1 is a factor of (3k + 2)x³ + (k - 1).

2x + 1 = 0 ⇒ x = -$\dfrac{1}{2}$

Since, 2x + 1 is a factor of (3k + 2)x³ + (k - 1)

∴ On substituting x = $-\dfrac{1}{2}$ in (3k + 2)x³ + (k - 1), remainder = 0.

$\Rightarrow (3k + 2)\Big(-\dfrac{1}{2}\Big)^3 + (k - 1) = 0 \\[1em] \Rightarrow \dfrac{-(3k + 2)}{8} + (k - 1) = 0 \\[1em] \Rightarrow \dfrac{-3k - 2 + 8k - 8}{8} = 0 \\[1em] \Rightarrow 5k - 10 = 0 \\[1em] \Rightarrow 5k = 10 \\[1em] \Rightarrow k = 2.$

Hence, k = 2.