I'm guessing not since it's not true for the integers (we have a negative and positive 'gcd', and we choose the positive one)
I'd appreciate some examples because I'm really new to fields and what-not.
Also, if it isn't unique, then how do we decide on which one to use?
On
Usually you get fractions. This method is equivalent to yours, I just like to write it out as continued fractions.
$$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) $$
$$ \left( 7 x^{4} + 6 x^{3} + 5 x^{2} + 4 x + 3 \right) $$
$$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) = \left( 7 x^{4} + 6 x^{3} + 5 x^{2} + 4 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 7 x + 8 }{ 49 } \right) } + \left( \frac{ 64 x^{3} + 128 x^{2} + 192 x + 270 }{ 49 } \right) $$ $$ \left( 7 x^{4} + 6 x^{3} + 5 x^{2} + 4 x + 3 \right) = \left( \frac{ 64 x^{3} + 128 x^{2} + 192 x + 270 }{ 49 } \right) \cdot \color{magenta}{ \left( \frac{ 343 x - 392 }{ 64 } \right) } + \left( \frac{ - 49 x + 1176 }{ 32 } \right) $$ $$ \left( \frac{ 64 x^{3} + 128 x^{2} + 192 x + 270 }{ 49 } \right) = \left( \frac{ - 49 x + 1176 }{ 32 } \right) \cdot \color{magenta}{ \left( \frac{ - 2048 x^{2} - 53248 x - 1284096 }{ 2401 } \right) } + \left( \frac{ 963342}{49 } \right) $$ $$ \left( \frac{ - 49 x + 1176 }{ 32 } \right) = \left( \frac{ 963342}{49 } \right) \cdot \color{magenta}{ \left( \frac{ - 2401 x + 57624 }{ 30826944 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( \frac{ 7 x + 8 }{ 49 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 7 x + 8 }{ 49 } \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 343 x - 392 }{ 64 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 49 x^{2} }{ 64 } \right) }{ \left( \frac{ 343 x - 392 }{ 64 } \right) } $$ $$ \color{magenta}{ \left( \frac{ - 2048 x^{2} - 53248 x - 1284096 }{ 2401 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 32 x^{4} - 832 x^{3} - 20064 x^{2} + 7 x + 8 }{ 49 } \right) }{ \left( \frac{ - 224 x^{3} - 5568 x^{2} - 133792 x + 160561 }{ 49 } \right) } $$ $$ \color{magenta}{ \left( \frac{ - 2401 x + 57624 }{ 30826944 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 49 x^{5} + 98 x^{4} + 147 x^{3} + 196 x^{2} + 245 x + 294 }{ 963342 } \right) }{ \left( \frac{ 343 x^{4} + 294 x^{3} + 245 x^{2} + 196 x + 147 }{ 963342 } \right) } $$ $$ \left( x^{5} + 2 x^{4} + 3 x^{3} + 4 x^{2} + 5 x + 6 \right) \left( \frac{ - 224 x^{3} - 5568 x^{2} - 133792 x + 160561 }{ 963342 } \right) - \left( 7 x^{4} + 6 x^{3} + 5 x^{2} + 4 x + 3 \right) \left( \frac{ - 32 x^{4} - 832 x^{3} - 20064 x^{2} + 7 x + 8 }{ 963342 } \right) = \left( 1 \right) $$
(the class of such $g$ are usually called the "associates" of $f$)
The $\Leftarrow$ direction is easy. The $\Rightarrow$ direction can be seen fairly straightforwardly from the definition of greatest common divisor. If $g$ is another greatest common divisor, then you have:
Combining these equations we see that $f = fuv$. If $f$ is nonzero, this implies $uv=1$, and thus $u$ is a unit. (the case where $f=0$ can be handeled separately)
Usually, for applications of greatest common divisors, any of them will work for your purpose. There is usually no need to single out a particular one to be "special" so that you should choose it. Sometimes there isn't even a good way of doing this.
Even if you do have a good reason to single out one to be special, it often doesn't matter how you do it; it only matters that for each particular class of associates you have a way to consistently pick out which associate is the special one.
For polynomials, however, the usual convention is to pick the monic polynomial to be the special one. (in the nonzero case, of course)