If $x^3+6x^2+4x+K$ is divisible by $(x+2)$, then the value of $K$ is

Let $f(x)=x^3+6x^2+4x+K$.
According to the Factor Theorem, since $f(x)$ is exactly divisible by $(x+2)$, we must have $f(-2)=0$.
Substitute $x=-2$ into the polynomial expression:

$$(-2{)}^3+6(-2{)}^2+4(-2)+K=0$$

$$-8+6(4)-8+K=0$$

$$-8+24-8+K=0$$

$$8+K=0⟹K=-8$$