How to compute the gcd of $x+a$ and $x+b$, where $a\neq b$? Having some trouble with a fairly basic field question.

Let $F$ be any field and $a \neq b$ where $a,b\in F$.
Find the greatest common divisor of $f(x)=x+a$ and $g(x)=x+b$, and also find polynomials $s(x)$ and $t(x)$ so that $$s(x)f(x)+t(x)g(x)= \gcd$$

Oddly the $s(x)f(x)+t(x)g(x)=\gcd$ is probably the easier part of this question but I am having a lot of trouble swallowing this notation; I'm only somewhat sure as to why $a=b$ is bad (I think it would allow $f(x)=g(x)$). Any help much appreciated.
 A: Consider polynomials over a field $F$. It will do no harm if for a while you think of $F$ as being the field of real numbers.   
Definition: Let $f(x)$ and $g(x)$ be two such polynomials, not both the $0$ polynomial. Then $d(x)$ is called a gcd of $f(x)$ and $g(x)$ if (i) $d(x)$ divides both $f(x)$ and $g(x)$, and (ii) for any polynomial $e(x)$ such that $e(x)$ divides both $f(x)$ and $g(x)$, we have that $e(x)$ divides $d(x)$.
This definition is analogous to the usual definition of gcd for integers. In fact, it is essentially identical to the usual definition in advanced courses.  
It is certainly not completely obvious that if $f(x)$ and $g(x)$ are polynomials, not both $0$, then $f(x)$ and $g(x) have a gcd. But they do. 
For example, $x-1$ is a gcd of $x^3-x$ and $x^2-3x+2$. But $-x+1$ is also a gcd of these two polynomials, as are $5x-5$ and  $\frac{1}{\pi}x-\frac{1}{\pi}$. These gcd's are all close relatives of each other.
And that is true in general. If $d(x)$ is a gcd of $f(x)$ and $g(x)$, then all the gcd's have shape $cd(x)$, where $c$ varies over non-zero elements of the field.
The polynomial $x+a$, $x+b$ where $a=b$: Look now at the polynomials $f(x)=x+a$  and $g(x)=x+a$. Certainly $x+a$ divides both $f(x)$ and $g(x)$. It is not had to show that if $e(x)$ divides $f(x)$ and $g(x)$, then $e(x)$ divides $x+a$. Thus $x+a$ is a gcd of $x+a$ and $x+a$.
The case $a\ne b$: Look now at $f(x)=x+a$ and $g(x)=x+b$, where $a\ne b$. We show that $1$ is a gcd of $f(x)$ and $g(x)$. (By the way, $17$ is also a gcd of these two polynomials.)
Certainly $1$ divides $x+a$ and $x+b$. Suppose that $e(x)$ divides both $x+a$ and $x+b$. We show that $e(x)$ divides $1$.
If $e(x)$ divides $x+a$ and $x+b$, then $e(x)$ divides their difference $(x-a)-(x-b)$, that is, $e(x)$ divides $a-b$. Let $a-b=e(x)q(x)$. Let $c=\frac{1}{a-b}$. Multiplying both sides by $c$, we obtain $e(x)(cq(x))=1$, so $e(x)$ divides $1$.
As you pointed out, finding polynomials $s(x)$ and $t(x)$ such that $s(x)(x+a)+t(x)(x+b)=1$ is easy. Just let $s(x)=\frac{1}{a-b}$ and $t(x)=-\frac{1}{a-b}$.  
Remark: In the case of polynomials over a field, there is (almost) an algorithm for computing the gcd of two polynomials. It is a variant of the usual Euclidean Algorithm for computing the gcd of two natural numbers. And it is quite algorithmic, with the caveat that during the execution, we need to do arithmetic on field elements, so that arithmetic has to be executable algorithmically.
A: The gcd of f and g is a polynomial. If a = b then f(x) = g(x) and f (or g) is the gcd trivially. You must come up with a polynomial that divides each of $x+a$ and $x+b$. A hint is that you are limited in the degree of the polynomial you must consider. Another is that the gcd of a and b might play a role in this.
If you have found the s and t then just write down the right hand side as the solution to part 1.
A: Hint $\rm\ \ 0\ne c\in F\:\Rightarrow\: gcd(f(x),f(x)\!-\!c) = 1\ $ since $\rm\ f(x)\, -\, (f(x)\!-\!c)\ =\ c$
