Examine whether $x+2$ is a factor of $x^{3}+3x^{2}+5x+6$ and of $2x+4$.

(N/A) The zero of $x+2$ is $-2$. Let $p(x) = x^{3}+3x^{2}+5x+6$ and $s(x) = 2x+4$.
For $p(x)$:
$p(-2) = (-2)^{3} + 3(-2)^{2} + 5(-2) + 6$
$p(-2) = -8 + 3(4) - 10 + 6$
$p(-2) = -8 + 12 - 10 + 6 = 0$.
Since $p(-2) = 0$,by the Factor Theorem,$x+2$ is a factor of $x^{3}+3x^{2}+5x+6$.
For $s(x)$:
$s(-2) = 2(-2) + 4$
$s(-2) = -4 + 4 = 0$.
Since $s(-2) = 0$,$x+2$ is a factor of $2x+4$. Alternatively,$2x+4 = 2(x+2)$,which clearly shows $x+2$ is a factor.