Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural. 
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$. 
Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.

Let us restate the value of $P$ first.
$P = (X+2)^m+(X+2)^{2m+3}+1^{2m+3} = (X+2)^m+(X+2)^{2m+3}+1 =$
$(X+2)^m+(X+2)^{2m+3}+(X+2)^0 = (X+2)^{3m+3} = (X+2)^{(m+1)^{3}}$
And, Let us restate the value of $Q$.
$Q = X^2+4X+4+X+3 = X^2+4X+4+X+2+1 = (X+2)^2+(X+2)^1+(X+2)^0$
Do you notice the relationship between $P$ and $Q$?
