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Let $f(x)=x^3-6x^2+x+4$ and $g(x)=x^5-6x+1\in \mathbb{Q}[x]$.

Find $\gcd(f,g)$ by using the Euclidean algorithm.(done)

Also use extended euclidean algorithm to find $s(x),t(x) \in Q[x]$ s.t. $\gcd(f,g)=fs+gt$. (not done)


Note that we are in $Q[x]$ land. $$\begin{aligned} x^5-6x+1 & =(x^3-6x^2+x+4)(x^2+6x+35)+(200x^2-65x-139) \\ x^3-6x^2+x+4 & =(200x^2-65x-139)\left (\frac{x}{200}-\frac{227}{8000} \right) + \left (\frac{239}{1600}x+\frac{447}{8000} \right) \\ 200x^2-65x-139 &=\left( \frac{-239}{1600}x+\frac{447}{8000} \right) \left (\frac{-3200}{239}x-\frac{375200}{57121}\right )+\frac{-7731076}{57121} \end{aligned}$$ So $\gcd(f(x),g(x))=1$

Setting up for reveres substitution

$$\begin{aligned} \frac{-7730176}{57121}& =(200x^2-65x-139) - \left (\frac{-239}{1600}x+\frac{447}{800} \right) \left ( \frac{-320000}{239}x-\frac{375200}{57121} \right) \\ \frac{-239x}{1600} +\frac{447}{8000} &= \left ( x^3-6x^2+x+4 \right) -\left ( \left (200x^2-65x -139 \right) \left ( \frac{x}{200}-\frac{227}{8000}\right) \right) \\ 200x^2-65x-139&= \left (\left (x^5-6x+1 \right ) -\left(x^3-6x^2+x+4\right)\left (x^2+6x+35 \right)\right) \end{aligned} $$ Call them 1,2,3 in decending order where 1 is constant=____

$3 \to 2$

$$ \frac{-239x}{1600} +\frac{447}{8000} = \left ( x^3-6x^2+x+4 \right) -\left ( \left (\left (x^5-6x+1 \right ) -\left(x^3-6x^2+x+4\right)\left (x^2+6x+35 \right)\right) \left ( \frac{x}{200}-\frac{227}{8000}\right) \right) $$

Running out of real state at this point but missing a step got $$\frac{-239x}{1600} +\frac{447}{8000} =(x^3-6x^2+x+4)\left (1-\left (x^2+6x+35\right) \left(\frac{x}{200}-\frac{227}{8000} \right) \right) -\left( x^5-6x+1 \right) \left (\frac{x}{200}-\frac{227}{8000}\right) $$

$2 \to 1$

$\frac{-7730176}{57121}=(200x^2-65x-139) - \left ( (x^3-6x^2+x+4)(1-(x^2+6x+35)(\frac{x}{200}-\frac{337}{800}))(\frac{-340000x}{239})-\frac{375,200}{571221} \right) + (-1)(x^5-6x+1)(\frac{x}{200}-\frac{227}{8000})(\frac{-320000x}{239}-\frac{375200}{57121}) $


Was able find the gcd but could no find the $sf+tg=1$

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The answer given by WA is really horrible:

$ s=\dfrac1{120784}(-5975 x^4-2235 x^3-4879 x^2-3139 x+30835) $

$ t=\dfrac1{120784}(5975 x^2-33615 x-2556) $

No wonder you're having trouble!

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