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Factor $x^3 + 2x + 3$ into irreducible polynomials in $\mathbb{Z} _5 [x]$

This polynomial has 2 zeros mod 5: x = 2 and x = 4. But these only give me a 2 degree polynomial $x^2 - 4$ and I don't know how to find the last one.

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Also you could note that the derivative is $3 x^2 + 2 = 3 (x^2 -1) = 3 (x-1)(x+1) = 3 (x - 1)(x - 4)$, so $4$ is a double root.

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  • $\begingroup$ wait a second here: if the derivative has a root mod n then the function has a root mod n? $\endgroup$ Feb 4 '14 at 7:13
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    $\begingroup$ If the polynomial and its derivative have a root in common, then that root is a multiple root of the polynomial. $\endgroup$ Feb 4 '14 at 7:43
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This polynomial has a double root. Which one? Besides trial and error, you could also use long division to divide your polynomial by $(x-2)(x-4)$.

$x^2-4$ is not relevant here; its roots are $2$ and $3$.

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  • $\begingroup$ I have no idea how i got $x^2 - 4$. It should be $x^2 + x - 2$. You're right, I could use long division, I'll do that now. could you also explain what it means to be a double root? I never really got a good grasp on that idea. $\endgroup$ Feb 4 '14 at 7:05
  • $\begingroup$ @terribleatmath A polynomial has a double root at $a$ if $(x-a)^2$ divides the polynomial. For example, $(x-2)^2=x^2-2x+4$ has a double root at $x=2$. $\endgroup$
    – Slade
    Feb 4 '14 at 7:07
  • $\begingroup$ thank you, that makes sense. it has a double root at x = 4. (trial and error :P) thank you! $\endgroup$ Feb 4 '14 at 7:11
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Hint $\ $ Let $\,r\,$ be the $3$rd root. By Vieta's Formulas the sum of the roots $= 0\ (=\,$ $-$coeff of $\,x^2),\,$ therefore $\, r + 2 + 4\equiv 0\pmod 5,\,$ so $\ r\equiv\, \ldots$

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