Find $\gcd(2x+7, x^2-2)=d(x)$ where all polynomials are in $\Bbb Q[x]$. So, I used the Euclidean Algorithm to solve for the GCD:
$x^2-2=(2x+7)(\frac12x-\frac74)+\frac{41}4$
$2x+7=\frac{41}4(\frac8{41}x+\frac{28}{41})+0$
$\therefore \gcd(x^2-2,2x+7)=\frac{41}4$

However, this is obviously not true, and I checked this multiple times. So, where did I go wrong?
 A: If the gcd turned out to be $x^3-5x+4$, that's the same as saying that it's $2x^3-10x+8$.  In this context, "greatest" in the phrase "greatest common divisor" means having the greatest degree among common divisors.  The two polynomials I give above are scalar multiples of each other and are therefore treated as equivalent (multiply the first by $2$ to get the second or multiply the second by $1/2$ to get the first).
Consider finding gcd's among integers $0,\pm1,\pm2,\pm3,\ldots\,{}$.  Is $\gcd(24,18)$ equal to $6$ or to $-6$?  You can get either of those numbers from the other by multiplying it by a member of the set $\{0,\pm1,\pm2,\pm3,\ldots\}$, namely by $-1$.  Therefore they're treated as equivalent.  But $6$ is not equivalent to $12$ since you cannot multiply $12$ by any member of $\{0,\pm1,\pm2,\pm3,\ldots\}$ and get $6$.
Among non-zero rational numbers, all are equivalent.
Sometimes one adopts the convention of regarding the monic polynomial among the possibilities as being the answer.  "Monic" means the leading coefficient is $1$, so in the example above it would be $x^3-5x+4$ and not $2$ times that.
Thus $41/4$ and $1$ are equivalent since you can get either from the other by multiplying it by some rational number, but the monic polynomial among them is $1$.
A: This is the same as 1 being the Gcd. In the rationals, $1\mid \frac{41}{4}$. 
A: More pertinently, you're indicating that the two polynomials are relatively prime. And this is true. 
One way of seeing this would be to actually pretend you were working in $\mathbb{R}[x]$. Then the first polynomial is $(x + \sqrt 2)(x - \sqrt 2)$, and the second is $2(x + \frac{7}{2})$. These clearly have no linear factors $(x - a)$ in common, and are thus clearly relatively prime.
