Let $$f(X) = 78X^3 + 174X^2 − 116 ∈ Z[X]$$

My question is to decide if $f(X)$ is irreducible in $Z[X], Q[X] and R[X]$

I have tried finding a prime number 29, and to fulfil the Eisenstein's Irreducibility Criterion

1) $29$ is a common factor of $-116$ and $174$ in $Z$;

2) $29^2 = 841$ is not a factor of $-116$ in $Z$

3) $29$ is not a factor of $78$.

So that its enough to show $f(X)$ is irreducible in $Q[X]$.

And for $R[X]$ there should be no root as a integers so its should be irreducible but are there any way i can prove there really not integers root? And i also getting stuck in how to check with $Z[X]$, are there anything needed from the above calculation or is it something new?

Thanks a lot!

  • $\begingroup$ Please add further tags, e.g. abstract-algebra $\endgroup$ Feb 11 '19 at 19:49
  • $\begingroup$ For $f(X)$ to be irreducible in $Bbb Q[X]$, it must have a linear factor of the form $ax+b$, where $a \vert 39$ and $b \vert 58$. $\endgroup$ Feb 11 '19 at 19:56
  • $\begingroup$ @Robert Not true, e.g. $\ (78x-1)(x-116).\,$ But OP already used Eisenstein to show irreducibility over $\,\Bbb Q.\ \ $ $\endgroup$ Feb 11 '19 at 20:11
  • $\begingroup$ Your question is confusing. You've shown there are no integers as roots because you've shown its irreducible over $\mathbb{Q}$ Are you asking how to show that it's irreducible over $\mathbb{R}$ $\endgroup$ Feb 11 '19 at 20:14
  • $\begingroup$ @Dione Hint: in $\,\Bbb Z[x]\,$ these are both reducible $\,2^2,\ 2x,\,$ being products of nonzero nonunits. See also here. $\endgroup$ Feb 11 '19 at 20:24

Every cubic polynomial with real coefficients has at least one real root and therefore it is reducible in $\mathbb{R}[x]$.

It is also reducible in $\mathbb{Z}[x]$, since it is equal to $2\times(39x^3+87x^2-58)$.

  • $\begingroup$ How are you concluding that it's reducible in $\mathbb{Z}[x]$ from that factorization? This cannot be correct because it's irreducible over $\mathbb{Q}$ $\endgroup$ Feb 11 '19 at 20:11
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    $\begingroup$ $2$ is a unit (i.e., a divisor of $1$) in $\Bbb Q$ but not in $\Bbb Z$. $\endgroup$ Feb 11 '19 at 22:00

One (other) way to show that it's irreducible over $\mathbb{R}$ is to notice that $f(0) < 0$ and $f(1) >0$. Hence, there must be a root between in $(0,1)$


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