I'm given $p(x)=x^3+4x^2+x-6, q(x)=x^5-6x+5$. I'm asked for the GCD in $\mathbb{Q}[x]$; which I know to be $x-1$ by factoring. However; writing $x-1$ as a combination of p and q (that is; finding $d_1(x),d_2(x) \in \mathbb{Q}[x]$ s.t. $qd_1(x)+pd_2(x)=x-1$)seems to require Euclid algorithm (this is how I would proceed if these where say, integers). This gets impossibly messy; I've tried 3 times now and none of my solutions checked out. Is there some more sophisticated trick or method I'm missing here?
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You can factor the first polynomial as $(x-1)(x+2)(x+3)$, and you can just check it manually. – Lynnx Sep 24 '20 at 10:47
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yes, this is true, the challge is writing x-1 as a linear combination of p and q... – Muselive Sep 24 '20 at 10:48
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1Does it answer your question? https://math.stackexchange.com/questions/2830421/is-this-method-to-calculate-gcd-of-two-polynomials-correct – Math Lover Sep 24 '20 at 11:04
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It is the exact same question asked earlier on the site. – Math Lover Sep 24 '20 at 11:05
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To keep the degrees low you could do write $1$ as a linear combination of $p/(x-1)$ and $q/(x-1)$, and then multiply both sides by $x-1$. – Jaap Scherphuis Sep 24 '20 at 11:07
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The Euclidean algorithm works perfectly fine. Two polynomial long divisions yield: \begin{array}{rrrrr} 1\times&(x^5-6x+5)&-&(x^2-4x+15)\times&(x^3+4x^2+x-6)&=&-50x^2-45x+95\\ 50\times&(x^3+4x^2+x-6)&-&(-x-\tfrac{31}{10})\times&(-50x^2-45x+95)&=&\tfrac{11}{2}(x-1).\\ \end{array} Substituting back then yields $$x-1=\frac{10x+31}{55}q+\frac{-10x^3+9x^2-26x+35}{55}p.$$
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