The zero of $x+2$ is $-2$ . Let $p(x)=x^{3}+3 x^{2}+5 x+6$ and $s(x)=2 x+4$

Then,                 $p(-2)=(-2)^{3}+3(-2)^{2}+5(-2)+6$

$=-8+12-10+6$

$=0$

So, by the Factor Theorem, $x+2$ is a factor of $x^{3}+3 x^{2}+5 x+6$

Again,                 $s(-2)=2(-2)+4=0$

So, $x+2$ is a factor of $2 x+4 .$ In fact, you can check this without applying the Factor Theorem, since $2 x+4=2(x+2)$.