I am finding the GCD of $a = x^3 + 11/3x^2 + 17/4x + 3/2$ and $b = 3x^2 + 22/3x + 17/4$ using the Euclidean algorithm. So I divide $a/b$ and get $q$ and $r$ such that $a = qb + r$. Then, according to Wikipedia, I set the new $a = b$, $b = r$, and repeat until $b = 0$, at which point the GCD of the original $a$ and $b$ will be the current $a$ value.
Iteration 1: $$a = x^3 + \dfrac{11}{3}x^2 + \dfrac{17}{4}x + \dfrac{3}{2}$$ $$b = 3x^2 + \dfrac{22}{3}x + \dfrac{17}{4}$$ $$x^3 + \dfrac{11}{3}x^2 + \dfrac{17}{4}x + \dfrac{3}{2} = \left(\dfrac{1}{3}x + \dfrac{11}{27}\right) \times \left(3x^2 + \dfrac{22}{3}x + \dfrac{17}{4}\right) - \dfrac{25}{162}x - \dfrac{25}{108}$$ Iteration 2: $$a = 3x^2 + \dfrac{22}{3}x + \dfrac{17}{4}$$ $$b = -\dfrac{25}{162}x - \dfrac{25}{108}$$ $$3x^2 + \dfrac{22}{3}x + \dfrac{17}{4} = \left(-\dfrac{486}{25}x - \dfrac{459}{25}\right) \times \left(-\dfrac{25}{162}x - \dfrac{25}{108}\right) + 0$$
Then I set $a = b = -\dfrac{25}{162}x - \dfrac{25}{108}$ and $b = r = 0$. Since $b$ is zero, the GCD is $a$.
Wolfram|Alpha gives the GCD as $1/6x + 1/4$. So what did I do wrong? (I'm sure it's simple, but I just can't figure out what I'm doing wrong...)
Note that: $$-\frac{25}{162}x - \frac{25}{108} = -\frac{25}{27}\left(\frac 1 6 x + \frac 1 4\right)$$
The GCD of 2 polynomials is determined up to a constant (aka a unit).