<s>[INST] Find the gcd of $f(x)$ and $g(x)$ Use the division algorithm to the find the gcd of $f(x)= 2x^4 + 5x^3 -5x -2$ and $g(x)= 2x^3 -3x^2 -2x$ in $\mathbb Q[x]$.
I used long division and found the following
$$f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right)$$
So I have found the polynomials $q(x)$ and $r(x)$ such that $$f(x) = q(x)g(x) + r(x)$$ where $\deg [r(x)] < deg[g(x)]$.
This is where I am stuck. How do I go about finding the GCD of $f$ and $g$? I know that it must be some monic ploynomial $d(x)$ 
 [/INST] They key property is that if $f=gq+r$, then $$\tag{1}\gcd(f,g)=\gcd(g,r).$$
So, you already found that 
$$
f(x) = (x+4)g(x) + \left( 14x^2 + 3x -2 \right).
$$
Now
$$
g(x)=\left(\frac17\,x-\frac{12}{49}\right)(14x^2+3x-2)+\left(-\frac{48}{49}x-\frac{24}{49}\right).
$$
And
$$
14x^2+3x-2=\left(-\frac{343}{24}x+\frac{49}{12}\right)\left(-\frac{48}{49}x-\frac{24}{49}\right).
$$
Then, using property $(1)$, 
$$
\gcd(f,g)=\gcd(g,14x^2+3x-2)=\gcd\left(14x^2+3x-2,-\frac{48}{49}x-\frac{24}{49}\right)=-\frac{48}{49}x-\frac{24}{49}
$$
(the last equality because $-\frac{48}{49}x-\frac{24}{49}$ is a factor of $14x^2+3x-2$). To get the monic version, we multiply by $-49/48$, to get
$$
\gcd(f,g)=x+\frac12.
$$
</s>