<s>[INST] Find the coefficients in Euclid's algorithm $a(x)=3x^4-4x^3-11x^2+4x+9$
$b(x)=3x^3+5x^2+x-1$
I find the greatest common divisor:
$$
\frac {3x^4-4x^3-11x^2+4x+9}{3x^3+5x^2+x-1}=x-3 \;\ \;\ mod= 3x^2+8x+6
$$
$$
\frac {3x^3+5x^2+x-1}{3x^2+8x+6}=x-1 \;\ \;\ mod= 3x+5
$$
$$
\frac {3x^2+8x+6}{3x+5}=x+1 \;\ \;\ mod= 1
$$
Greatest common divisor $= 1$
I am looking for coefficients using the extended Euclidean algorithm, table: 
$3x^4-4x^3-11x^2+4x+9 \;\ \;\ \;\ \;\ 1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 0$
$3x^3+5x^2+x-1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 0 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 1$
$3x^2+8x+6 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ 1 \;\ \;\ \;\ \;\ -x+3$
$3x+5 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ -x+1 \;\ \;\ \;\ \;\ \;\ \;\ ?$
$1 \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ $?$ \;\ \;\ \;\ \;\ \;\ \;\ \;\  $?$ $
Checking:
$ $a(x)$(3x^4-4x^3-11x^2+4x+9)+$b(x)$(3x^3+5x^2+x-1)=1$
 [/INST] In general, $a = bq+r$, so $1 \cdot a + (-q) \cdot b = r$.  To continue the calculations, we introduce the notation $(\alpha,\beta)$ to denote $\alpha \cdot a + \beta \cdot b$, so that it's convenient to calculate the linear combinations of $a$ and $b$.  Therefore, $r$ becomes $(1,-q)$.
In the Extended Euclidean Algorithm, since $r_{i+1} = r_{i-1} - q_ir_i$, we have
$$\begin{aligned}
(\alpha_{i+1},\beta_{i+1}) &= (\alpha_{i-1},\beta_{i-1}) - q_i(\alpha_i,\beta_i) \\
&= (\alpha_{i-1}-q_i\alpha_i,\beta_{i-1}-q_i\beta_i).
\end{aligned}$$
The follow table illustrates the calculations for $\alpha_{i+1}$ and $\beta_{i+1}$ from the last two rows.
\begin{array}{|r|r|r|r|}\hline
\text{index} & \text{quotient } q_{i-1} & \hfill \text{remainder } r_i \hfill & \alpha_i & \beta_i \\ \hline
i-1 & q_{i-2} & r_{i-1} & \color{red}{\alpha_{i-1}} & \color{red}{\beta_{i-1}} \\ \hline
i & q_{i-1} & r_i & \color{blue}{\alpha_i} & \color{blue}{\beta_i} \\ \hline
i+1 & \color{blue}{q_i} & r_{i+1} & \color{red}{\alpha_{i-1}}-\color{blue}{q_i\alpha_i} & \color{red}{\beta_{i-1}}-\color{blue}{q_i\beta_i} \\ \hline
\end{array}
From your (correct) calculations for the Euclidean Algorithm, the first few rows of the following table are fixed.
\begin{array}{|r|r|r|r|}\hline
\text{index} & \text{quotient } q_{i-1} & \hfill \text{remainder } r_i \hfill & \alpha_i & \beta_i \\ \hline
1 & & 3x^4-4x^3-11x^2+4x+9 & 1 & 0 \\ \hline
2 & & 3x^3+5x^2+x-1 & 0 & 1 \\ \hline
3 & x-3 & 3x^2+8x+6 & 1 & -x+3 \\ \hline
4 & x-1 & 3x + 5 & -x+1 & x^2-4x+4\\ \hline
5 & x+1 & 1 & x^2 & -x^3+3x^2-x-1\\ \hline
\end{array}
Checking:
$$\begin{aligned}
& x^2 (3x^4-4x^3-11x^2+4x+9) + (-x^3+3x^2-x-1)(3x^3+5x^2+x-1) \\
&= 3x^6-4x^5-11x^4+4x^3+9x^2 + (-3x^6+4x^5+11x^4-4x^3-9x^2+1) \\
&= 1
\end{aligned}$$
</s>