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Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^3+6x^2+8x+4$

Dear Community, it took me some hours to think about it and i cant come to a good solution, i would be so happy if someone can show me how to do this right...^^

So this is my attempt:

$p(x) = x^3+6x^2+8x+4$

substitute with $(x-1)$ gives:

$p'(x-1) = x^3+3x^2-x+1$

modulo $2$ gives:

$p'(x-1) = x^3+x^2+x+1$

now substitute with $(x+1)$

$p''(x+1) = x^3+4x^2+6x+4$

now we se $x=-2$ we get $(x+2)(x^2+2x+2)$ so its reducible?

Vek
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    Reducible mod $2$ does not imply reducible over $\Bbb Q$. Note that the polynomial is of degree $3$, so it is reducible iff it has a root in $\Bbb Q$. (Hint: Rational root theorem) – leoli1 Jan 12 '21 at 09:34
  • Your work is nonsense, but if you are claiming that $p(x)=(x+2)(x^2+2x+2)$, then presumably you should able to mutiply two polynomials and see for yourself if that's the case. –  Jan 12 '21 at 09:36
  • @leoli1 Do you mean p(x) mod 2 gives $x^3$, $x^3$ is usually reducible, but with using rational root theorem in $Q$, we see there is no root in $Q$, so its irreducible? – Vek Jan 12 '21 at 10:21
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    That is correct. The polynomial is irreducible over $\Bbb Q$ but not modulo $2$. – leoli1 Jan 12 '21 at 10:23

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