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I was reading this inequality question. Here is part of the text that I have question about:

$$2t^8-t^7-8t^6+t^5+15t^4-4t^3-11t^2+4t+4>0.$$ Using Maple, I got $$(2t^2-t-2)(t^2-t-1)(t^4+t^3-t^2-t+2)>0.$$

And by factoring we can solve the inequality. but I wonder is it possible to factor the polynomial $$2t^8-t^7-8t^6+t^5+15t^4-4t^3-11t^2+4t+4$$ By hand without using any math software? By the result of the factoring we can see the roots of polynomial are $\frac{1\pm\sqrt{17}}{4}$ and $\frac{1\pm\sqrt5}{2}$, So we couldn't guess a number for example an integer like $a$ and see it makes the equation zero to have the factor $t-a$, also I can't see any obvious way to factor the polynomial so I wonder how we can factor it by hand.

Etemon
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    If you have the inspiration to try $t -1/t = z$ you get a polynomial in $z$ with $2$ rational double roots which correspond to the quadratic factors in $t$. – dxiv Jun 17 '21 at 00:15
  • @dxiv what is the inspiration for that, though? I've seen the technique in AoPS's Intermediate Algebra book but I forgot what makes one choose to do that. – OlympusHero Jun 20 '21 at 22:30
  • @OlympusHero You may think at reciprocal polynomials, but this not one of them. There isn't really any systematic way to get (or justify) such "inspiration" for this one in particular. It's obvious, of course, if you know the factorization already. Otherwise, it's just a case of grasping at straws, and noticing perhaps that the symmetric coefficients are "close" enough that it may be worth trying a $t \pm 1/t$ substitution. – dxiv Jun 23 '21 at 08:29
  • If you really had to do this by hand, substituting in $t=10$ gives $182244944$ for the expression, which when prime factorized (takes some work, but better than in variables) is divisible by $89$, this is suspiciously close to $100-10+1$=$t^2-t+1$ so try it? – xrider1000 Nov 09 '21 at 13:46

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