<s>[INST] Given $n \in \Bbb Z$, determine $\gcd(3n^2 + 7n + 4, n + 2)$. I factored $3n^2+7n+4$ to $(3n+4)(n+1)$ and because there isn't a common factor of those and $n+2$ I said that the gcd is $1$, but is there any othere way to go about it that would come up with a gcd that isn't $1$?
 [/INST] We have
$$3n^2+7n+4 = (n+2)(3n+1) + 2$$
Hence, if $d$ is a common divisor of $3n^2+7n+4$ and $n+2$, then $d$ must divide $2$. Hence, $\gcd(3n^2+7n+4,n+2) = 1 \text{ or }2$. $3n^2+7n+4$ is always even. Further, $n+2$ has the same parity as $n$.
Hence, if $n$ is odd, $$\gcd(3n^2+7n+4,n+2) = 1$$
while if $n$ is even $$\gcd(3n^2+7n+4,n+2) = 2$$
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