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