<s>[INST] Polynomial with "symmetrical" coefficients, and $\gcd(p,p')$ Consider the following polynomial:
$p(x) = x^6+x^5+2x^4+x^3+2x^2+x+1$. I would like to calculate $\gcd(p,p')$. 
With Euclid's GCD algorithm, the calculation becomes very complicated, very quickly. I was wondering if the fact that the coefficient vector is "palindromic" could be of some use, but I didn't notice any interesting pattern while I was doing the calculation. Is there anything I'm missing?
 [/INST] $p(x)=x^6 + x^5 +  x^4 +x^4+ x^3 + x^2 +x^2+ x + 1$
$p(x)=x^4(x^2+x+1)+x^2(x^2+x+1)+1(x^2+x+1)=(x^2+x+1)(x^4+x^2+1)=(x^2+x+1)[(x^4+2x^2+1)-x^2]=(x^2+x+1)[(x^2+1)^2-x^2]=(x^2+x+1)(x^2+x+1)(x^2-x+1)=(x^2+x+1)^2(x^2-x+1)$
$p'(x)=2(x^2+x+1)(2x+1)(x^2-x+1)+(x^2+x+1)^2(2x-1)=(x^2+x+1)(6 x^3-x^2+3 x+1)$
Let $q(x)=6x^3-x^2+3x+1$ 
$q(\pm1)\ne 0;\;q\left(\pm\frac12\right)\ne 0;\;q\left(\pm\frac13\right)\ne 0;\;q\left(\pm\frac16\right)\ne 0$
$q(x)$ has no rational roots therefore it is irreducible in $\mathbb{Q}$
$\text{GCD}(p(x),p'(x))=x^2+x+1$
Hope this helps
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