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this question follows one from yesterday that got deleted because it was a duplicate.

The problem was about solving this equation (in $ℚ[x]$) :

$$f(x)(2x^3 + 3x^2 + 7x + 1) + g(x)(5x^4 + x + 1) = x + 3$$

Members of the community answered yesterday that the gcd is 1 which implies that the equation is solvable.

Based on their advices I tried to use Extended Euclidian Algo for polynomials to find the Bezout coeff such that $a(x)(2x^3+3x^2+7x+1)+b(x)(5x^4+x+1) = 1$.

But after very long calculations, I couldn't find 1 as the last reminder but $\frac{26684375}{261662976} \simeq 0.102$. I hope it's a miscalculation, but I'm lost.

Can you tell me how members that answered me yesterday found 1 as the gcd that quickly ?

Thanks in advance.enter image description here

enter image description here

AANICR
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  • Please do not post images of work as it is bad for accessibility. Instead, use Latex. – codeing_monkey Dec 18 '23 at 14:55
  • Is the following question the "one from yesterday": Equation in ring $Q[x]$? That question is not deleted. – peterwhy Dec 18 '23 at 15:48
  • @peterwhy Yes it's this one. I thought it was deleted sorry. – AANICR Dec 18 '23 at 16:05
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    As explained in the linked dupe, gcds that are constant $,c\in \Bbb Q,$ are normalized to $1$ by convention. Generally we can ignore nonzero constant (= unit) factors when computing polynomial gcds (over a field) since $,f\mid cg\iff f\mid g.\ \ $ – Bill Dubuque Dec 18 '23 at 17:53
  • Thanks @BillDubuque ! Does it mean that i can ignore the constant and act as if I obtained 1 ? Or do I have to multiply my whole last equation by the inverse of the constant I obtained (on both sides of the equation) ? – AANICR Dec 19 '23 at 13:24
  • $a,f+b,g = c\iff c^{-1}a,f + c^{-1}b,g = 1$. By convention we refer to $1$ as "the" gcd of $f,g.,$ More generally we normalize the gcd so that it is monic (lead coef $c= 1),,$ see the linked dupe. – Bill Dubuque Dec 19 '23 at 16:16

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