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I know that $a$ is root of $p$ iff $p(a)=p'(a)=0$

$p(x)= x^5 + x + 1$

$p'(x)= 5x^4 + 1$ then $p'(a)=0$ iff $5a^4 + 1 = 0$ iff $a^4=-1/5$

Now, $p(a)= a^5 + a + 1 = a. a^4 + a + 1 = (-1/5) a + a + 1 = (4/5) a + 1 = 0$

then, $a = 1 - 5/4$ so $a = -1/4$. But $(-1/4)^4 ≠ -1/5$ so there is no solution. So $p$ does not have multiple roots.

So $p$ has all its simple roots.

Is my exercise resolution correct? Are the steps well explained and justified?

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    $\begingroup$ Where'd the $a^4$ go? $\endgroup$ – Badr B Jun 7 '18 at 21:39
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    $\begingroup$ I suppose you mean $a$ is a multiple root if and only if … $\endgroup$ – Bernard Jun 7 '18 at 21:44
  • $\begingroup$ "... then $a=1-5/4$" is not correct. $\endgroup$ – A.Γ. Jun 7 '18 at 21:45
  • $\begingroup$ @BadrB I wrote badly that step. The correct is $a^4=-1/5$. I corrected it. $\endgroup$ – Ayesca Jun 7 '18 at 21:56
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    $\begingroup$ From $(4/5)a+1=0$ you should get $a=-5/4$, not $a=-1/4$. But your reasoning is otherwise correct, since $(-5/4)^4\not=-1/5$ either. $\endgroup$ – Barry Cipra Jun 7 '18 at 21:58
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You made a minor mistake when you wrote that $5a^4+1=0\iff a=-1/5$. You should have written $a^4=-1/5$ instead.

The rest is correct, assuming that you are working over a field with characteristic $0$.

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  • $\begingroup$ You're right, I wrote badly that step. Thank you very much, I corrected it. $\endgroup$ – Ayesca Jun 7 '18 at 21:53
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    $\begingroup$ @Ayesca I'm glad I could help. $\endgroup$ – José Carlos Santos Jun 7 '18 at 21:57
  • $\begingroup$ @JoséCarlosSantos, you overlooked a second minor error. See my comment below the OP. $\endgroup$ – Barry Cipra Jun 7 '18 at 22:03
  • $\begingroup$ @BarryCipra You're right: I missed that. $\endgroup$ – José Carlos Santos Jun 7 '18 at 22:04
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Consider $f(x)=x^5+x+1$ and $f'(x)=5x^4+1$ like you said. Since $f'(x)$ is always positive, we can conclude that $f(x)$ is always increasing. This means that the function only crosses the x-axis once (since $f(x)$ is of odd degree). Therefore we can conclude that $f(x)$ has only one real root, and since the function is always increasing, it must be a simple root.

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    $\begingroup$ The OP wants to show that all the roots are simple, not just the real roots. $\endgroup$ – Barry Cipra Jun 7 '18 at 22:04
  • $\begingroup$ @BarryCipra Ohh I see. Thanks for the pointing that out. $\endgroup$ – Badr B Jun 7 '18 at 22:13

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