Finding Linear Combination of Polynomials I am stuck on a question involving finding the greatest common divisor of polynomials and then solving to find the linear combination of them yielding the greatest common divisor. My work thus far is as follows:
Define: $f(x) = x^3-3x+3$, $g(x) = x^2 - 4$. 
Execute the Euclidean Algorithm for Polynomials as follows:
$x^3-3x+3 = x(x^2 -4)+(x+3)$
$x^2-4 = (x-3)(x+3) +5$
$x+3 = \frac x5(5) +3$
$5 = 1(3) +2$
$3 = 1(2) +1$
$2 = 1(2) +0$
So we know that $\gcd(f(x),g(x)) = 1$. Now, my objective is to find polynomials $s(x), t(x)$ such that $f(x)s(x) + g(x)t(x) = 1$. 
I have tried using back substitution, a method involving matrix multiplication of coefficients, and a method using continuous fractions. None of these have worked- all returning incorrect answers, two of which were the same. Does anyone have any suggestions as to where I am going wrong, or a simpler way to finish this computation?
Note: it is not specified in my book which field these polynomials are defined over. I am assuming the reals.
 A: The first two lines are fine. Then you are finished, though you may want to write $x+3=\frac{x+3}{5}\cdot 5+0$. 
Work backwards from the second line. We get $5=x^2-4 -(x-3)(x+3)$. But from the first line we have $x+3=x^3-3x+3-x(x^2-4)$. Substituting we get
$$5=x^2-4-(x-3)(x^3-3x+3-x(x^2-4)).$$
This simplifies to 
$$5=(3-x)(x^3-3x+3)+(x^2-3x+1)(x^2-4).$$
If you want a $1$ on the left-hand side, multiply through by $\frac{1}{5}$. If you are working in a field that does not contain the rationals (but is not of characteristic $5$), instead of multiplying by $\frac{1}{5}$, multiply by the inverse of $5$.
A: Just use the bog-standard extended Euclidean algorithm, as given by Blankinship (a terse description here). If you look closely, it assumes only that the elements manipulated have a division algorithm, i.e., form an Euclidean domain.
Polynomials over any field form an Euclidean domain, so the exact field is irrelevant.
A: We chose polynomials $f(x)$ and $g(x)$ such that $$f(x)(x^2-4) + g(x)(x^3-3x+3)=1$$ and for this it is necessary the equality of degrees.
So $d(f)+2=d(g)+3\iff d(f)=d(g)+1$. 
Chose the simplest possibilities
$f(x)=ax^2+bx+c$ and $g(x)=dx+e$ and calculate the coefficients.It follows
$(a+d)x^4+(b+e)x^3+(-4a+c-3d)x^2+(-4b+3d-3e)x+(-4c+3e)=1$
Solving the resulting five equations equalizing coefficients we get
$f(x)=\frac {x^2-3x+1}{5}$ and $g(x)=\frac {-x+3}{5}$ which gives the identity
$$(x^2-3x+1)(x^2-4)+ (-x+3)(x^3-3x+3)=5$$
