How do I calculate $\operatorname{Cov}(X + 3, X\cdot Z)$ if I know that

$$ f\_X(x) = \begin{cases} 4x^3 &, \text{if} \ 0 < x < 1 \\
0 &, \text{otherwise} \end{cases}$$ where $Z \sim N(1,2)$ and $X$ and $Z$ are independent?

I started with $\operatorname{Cov}(X+3,X\cdot Z) = \operatorname{Cov}(X,X \cdot Z)$ and thought maybe I should use that $\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y]$ where I would set $Y = X \cdot Z$ but that would end up giving $0$ and I know the answer is $2/75$.

What property can I use?

Thanks.