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Find all c $ \in Q\ if\ any\ s.t. f(c)=0$ Where i) $f(x)=4x^4-2x^3-2x^2-x-1$ ii) $f(x)=8x^3-6x-1$

Okay so what I have been trying is the rational root test. In i the possible roots are: +-$({1,1/2, 1/4 })$ but none of them are roots. Then I can see that Eisenstein criterion is not working as well. So I tried to see if it is irreducible in $Z_{p}\ with \ Z_{2}$. Hence if this is the case I could conclude that it is irreducible in $Q_[x]$, but it is reducible since in $Z_{2} \ i)=x+1$ which is reducible in $Z_{2}$... so now I am quite stuck my approach for ii) was the same but also got stuck

  • What's the question? Is it reducibility or roots? In any case, you can't use $p=2$ for Eisenstein (for the first one) as the lead coefficient is $0\pmod 2$. – lulu Jan 08 '18 at 11:44
  • I have to find f(c)=0 if any –  Jan 08 '18 at 11:47
  • Reducibility should not confused with solutions, I think the OP is trying to find roots. – Landuros Jan 08 '18 at 11:47
  • Then why are you talking about Eisenstein? The rational root theorem is all you need. – lulu Jan 08 '18 at 11:47
  • Please edit your post for clarity. The header clearly address reducibility, as does much of the body. That's a much harder question then just finding roots. (worth noting: for cubics, finding rational roots is the same as deciding reducibility, but not for higher degree). – lulu Jan 08 '18 at 11:49
  • Okay the edit is done now. –  Jan 08 '18 at 11:50
  • So what you're saying is that I can just use rational root test, and then I can see that none of the possible roots are roots? –  Jan 08 '18 at 11:50
  • Exactly! That's what the rational root theorem says. – lulu Jan 08 '18 at 11:51
  • Okay, I just got confused with irreducibility and roots. I was thinking that if you have roots, then it must be reducible. But if there is no roots then it is irreducible –  Jan 08 '18 at 11:52
  • @Djhoe Specifically, there are no integer roots. – Landuros Jan 08 '18 at 11:56
  • ok thanks so for the ii) +-(1,1/2,1/4,1/8) are no roots $ \implies $ no roots exists as well –  Jan 08 '18 at 12:02
  • BTW, the roots of $8x^3-6x-1$ are $\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$, all irrational. See https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root – lhf Jan 08 '18 at 12:05

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