Find a linear representation of the greatest common divisor in terms of the polynomials $f(x)$ and $g(x)$

Question

There are two functions: $$ f(x)=x^5+x^4-x^3-2x-1,\\ g(x)=3x^4-2x^3+x^2-2x-2 $$ And I've already found that their GCD equals $r_2=\frac{-9x^2}{4}-\frac{9}{4}$. Here are some other calculations performed by me: $$ f(x)=\left(\frac{x}{3}+\frac{5}{9}\right)g(x)+r_1\\ r_1=\left(\frac{8x}{81}-\frac{4}{81}\right)r_2\\ g(x)=\left(\frac{-27x}{2}+\frac{9}{4}\right)r_1+r_2 $$ I need to linearly express $r_2$ in terms of $f(x)$ and $g(x)$. Is it correct to write the following? $$ r_1=\left(\frac{8x-4}{81}\right)r_2=f(x)-\left(\frac{x}{3}+\frac{5}{9}\right)g(x)\\ r_2=\left(\frac{81}{8x-4}\right)f(x)-\left(\frac{81}{8x-4}\right)\left(\frac{3x+5}{9}\right)g(x)\\ r_2=\left(\frac{81}{8x-4}\right)f(x)-\left(\frac{27}{8}(8x-4)+\frac{117}{2}\right)g(x) $$

Answer 1

Suppose you are applying the Euclidean algorithm:

$$f=q_1g+r_1$$

$$g=q_2r_1+r_2$$

$$r_1=q_3r_2+r_3$$

Then you can work backwards

$$r_3=r_1-q_3r_2=(f-q_1g)-q_3(g-q_2r_1)=f-q_1g-q_3g+q_2q_3(f-q_1g)=$$$$=(1+q_2q_3)f-(q_1+q_3+q_1q_2q_3)g$$

I hope that is sufficient clue as to how to go about this.