Struggling to see how the $\operatorname{gcd}(f,g)=1$ I'm working with two polynomials:
$$
\begin{eqnarray}
f(x) &=& 2t^{3} - 3t^{2} + 6 \\
g(x) &=& 6t^{2} - 6t
\end{eqnarray}
$$
and I'm doing the long division to find the $\operatorname{gcd}(f,g)$ using the Euclidean algorithm. I know from my notes that $\operatorname{gcd}(f,g)=1$, and after doing the process of long division I find that $2t^3-3t^2+6 = (t/3 - 1/6)(6t^2 - 6t) + (-t+6)$. I can't figure out how repeating the long division would imply I have a $\operatorname{gcd}$ of $1$. Any help with this would be much appreciated
 A: When I talk about polynomials below, you can assume that the coefficients are complex numbers, but everything I'm going to say is going to work just as well if the coefficients are elements of a field.
First, we have the following fact: for any polynomials $f(t)$ and $g(t)$, if $g(t)\ne0$, then there exist unique polynomials, $q(t)$ and $r(t)$ such that 
$$f(t)=g(t)\cdot q(t)+r(t)$$
and either $r(t)=0$ or $\deg r(t)<\deg g(t)$. We find $q(t)$ and $r(t)$ by doing long division. If we divide $f(t)$ by $g(t)$, then $q(t)$ will be the quotient, and $r(t)$ will be the remainder.
The idea behind the Euclidean Algorithm is this: $GCD(f(t),g(t))=GCD(g(t),r(t))$.
You mentioned that you were given that
$$f(t)=2t^3-3t^2+6$$
$$g(t)=6t^2-6t$$
and that you divided and got that 
$$2t^3-3t^2+6=(6t^2-6t)\cdot\left(\frac{1}{3}t-\frac{1}{6}\right)+(-t+6).$$
Then $q(t)=\frac{1}{3}t-\frac{1}{6}$, and $r(t)=-t+6$. So the Euclidean Algorithm tells us that
$$GCD(2t^3-3t^2+6,6t^2-6t)=GCD(6t^2-6t,-t+6).$$
The GCD on the right will be easier to find since we're dealing with smaller degree polynomials. If we divide again, we can show that this GCD will be equal to a GCD that is even easier to find.
Dividing $6t^2-6t$ by $-t+6$, we get that
$$6t^2-6t=(-t+6)\cdot(-6t-30)+180.$$
Now we have that
$$GCD(6t^2-6t,-t+6)=GCD(-t+6,180).$$
Hence we have that
$$\begin{align*}
GCD(2t^3-3t^2+6,6t^2-6t)&=GCD(6t^2-6t,-t+6)\\
&=GCD(-t+6,180)\\
&=1\end{align*}$$
The fact that $GCD(-t+6,180)=1$ is because of the way GCD is defined. If $a(t)$ and $b(t)$ are polynomials that are not both zero, then $GCD(a(t),b(t)$ is the monic polynomial of largest degree that divides both $a(t)$ and $b(t)$. A polynomials is monic iff its leading coefficient is $1$. The only monic polynomial that divides $-t+6$ and $180$ is $1$. Hence $GCD(-t+6,180)=1$.
