If $\alpha$ and $\beta$ are the zeroes of polynomial $3x^2 + 6x + k$ such that $\alpha + \beta + \alpha\beta = -\frac{2}{3}$, then the value of $k$ is:

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Step 1: For polynomial $3x^2 + 6x + k$: $\alpha + \beta = \frac{-6}{3} = -2$ and $\alpha\beta = \frac{k}{3}$. Given $\alpha + \beta + \alpha\beta = -\frac{2}{3}$, so $-2 + \frac{k}{3} = -\frac{2}{3} \Rightarrow \frac{k}{3} = \frac{4}{3} \Rightarrow k = 4$

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