Let $p(x)=a x^{3}+x^{2}-2 x+4 a-9$

As $(x+1)$ is a factor of $p(x)$ $\therefore \quad p(-1)=0$ [By factor theorem]

$\Rightarrow \quad a(-1)^{3}+(-1)^{2}-2(-1)+4 a-9=0$

$\Rightarrow \quad a(-1)+1+2+4 a-9=0$

$\Rightarrow \quad-a+4 a-6=0$

$\Rightarrow \quad 3 a-6=0 \Rightarrow 3 a=6 \Rightarrow a=2$