<s>[INST] Greatest common divisor for $x+1$ and $2x+1$? I was just wondering, what would the gcd of $x+1$ and $2x+1$ be? 
I know that $x+1$ is not even or odd, whereas, $2x+1$ is odd. Is the gcd $1$?
 [/INST] If $d(x)|x+1$ and $d(x)|2x+1$ then $d(x)|(2x+1)-(x+1)= x$.  And so $d(x)|x$ and $d(x)|x+1$ so $d(x)|(x+1) -x = 1$ so $d(x)|1$. And assuming we are talking $\mathbb Z[x]$ this is as far as we go and $d(x) =1$ is the only (positive) common divisor of $2x+1$ and $x+1$.  So the greatest common divisor of $2x + 1$ and $x+1$ is $1$.
In general you have the rule $\gcd(a,b) = \gcd(a \pm k b, b)$ so $\gcd(2x+1, x+1) = \gcd(x, x+1) = \gcd(x, 1) = 1$.
Euclid's algorithm confirm:.
$(2x+1) = (x+1) + (x)$
$(x+ 1) = (x) + (1)$
$(x) = (1)\cdot x + 0$ so $\gcd(2x+1, x) = 1$.
As does Bezout's Lemma:
$2\cdot(x+1) - 1\cdot(2x+1) = 1$
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