Greatest common divisor of polynomials over $\mathbb{Q}$

Question

I have two polynomials: $f: x^3 + 2x^2 - 2x -1$ and $g: x^3 - 4x^2 + x + 2$. I have to do two things: find $gcd(f,g)$ and find polynomials $a,b$ such as: $gcd(f,g) = a \cdot f + b \cdot v$. I have guessed their greatest common divisor: $(x-1)$, but I did it by looking for roots of both polynomials, and now I am stuck. How do I find the greatest common divisor using the Euclid algorithm? I started with $f(x) = g(x) + 3(2x^2 - x - 1)$, but then things go nuts, and I can't use Bézout's identity to bring it all back to $gcd(f,g) = a \cdot f + b \cdot v$.

Answer 1

The Extended Euclidean Algorithm described here also works for polynomials over any field. For example, below is a computation of $\,\gcd(x^4+x+1,x^2+1)\,$ over $\,\Bbb F_2 = \,$ integers mod $2$.

$$\begin{array}{cccc} & x^4\!+x+1 & 1 & 0\\ & x^2\!+1 & 0 & 1\\ &x^2\!+x+1 & 1 & x^2\\ & x & 1 & x^2\!+1\\ & 1 & x & x^3\!+x+1\\ \end{array}$$

where each row $\ a\ \ b\ \ c\ $ means that $\ a = b(x^4\!+x+1) + c(x^2\!+1).\ $ Hence, by the final row

$$ 1\, =\, x(x^4\!+x+1) + (x^3\!+x+1)(x^2\!+1)$$

which, finally, implies that $\ (x^3\!+x+1)(x^2\!+1) = 1\,$ in $\,\Bbb F_2[x]/(x^4\!+x+1).$

Answer 2

Hints: $$x³+2x²-2x-1= (x-1)(3x+x²+1)$$ and $$ x³-4x²+x+2= (x-1)(-3x+x²-2) $$