lf $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+4{\mathrm{x}}^2+\mathrm{\lambda x}+1$is a monotonically decreasing function of x in the largest possible interval (-2, -2/3). Then

Here, ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\le 0$
or    $3{\mathrm{x}}^2+8\mathrm{x}+\mathrm{\lambda }\le 0 \mathrm{\forall }\mathrm{x}\in \left(-2,-\frac{2}{3}\right)$
Then situations form f '(x) are as follows: 
 

Given that /(*) decreases in the largest possible interval $\left(-2,-\frac{2}{3}\right)$.

Then $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$must have roots -2 and -2/3.

Thus, product roots $=(-2)\left(-\frac{2}{3}\right)=\frac{\mathrm{\lambda }}{3}\text{ or }\mathrm{\lambda }=4$