Using the division algorithm, find the greatest common divisor of $x^4+x+1$ and $x^3+2$ in $\mathbb{Q}[x]$ and express it as a linear combination of these polynomials.
So far the division has given me that:
\begin{align*} x^4+x+1&=x(x^3+2)+(1-x)\\ x^3+2&=(-x^2-x-1)(1-x)+3\\ 1-x&=-\frac{1}{3}x+\frac{1}{3} \end{align*} and of course that $\gcd\left(f(x),\,g(x)\right)=1$.
I'm not entirely sure how to proceed or how to represent these as linear combinations.
Once you get to $x^3+2 = (-x^2-x-1)(1-x) + 3$, you are done because the remainder is a unit. Or you can notice that in your last step, the remainder is $0$.
Now you can back-substitute: $$\begin{align*} 3 &= (x^3+2) - (-x^2-x-1)(1-x)\\ &= (x^3+2) - (-x^2-x-1)\Bigl( (x^4+x+1) - x(x^3+2)\Bigr)\\ &= (1 +x(-x^2-x-1))(x^3+2) - (x^3-x-1)(x^4+x+1)\\ &= (-x^3-x^2-x+1)(x^3+2) + (1+x-x^3)(x^4+x+1). \end{align*}$$ Dividing through by $3$ gives you an expression for $1$ in the form $p(x)(x^3+2) + q(x)(x^4+x+1)$, which is what you want.
There are other ways than back-substitution, if you keep track of what you are doing along the way. Essentially, the exact same methods as for integers, but with polynomials and long division.
