Decide if f(X) is irreducible in the following rings

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!

• Please add further tags, e.g. abstract-algebra Feb 11 '19 at 19:49
• 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 $b \vert 58$. Feb 11 '19 at 19:56
• @Robert Not true, e.g. $\ (78x-1)(x-116).\,$ But OP already used Eisenstein to show irreducibility over $\,\Bbb Q.\ \$ Feb 11 '19 at 20:11
• 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 $\mathbb{R}$ Feb 11 '19 at 20:14
• @Dione Hint: in $\,\Bbb Z[x]\,$ these are both reducible $\,2^2,\ 2x,\,$ being products of nonzero nonunits. See also here. Feb 11 '19 at 20:24

2 Answers

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 $$2\times(39x^3+87x^2-58)$$.

• How are you concluding that it's reducible in $\mathbb{Z}[x]$ from that factorization? This cannot be correct because it's irreducible over $\mathbb{Q}$ Feb 11 '19 at 20:11
• $2$ is a unit (i.e., a divisor of $1$) in $\Bbb Q$ but not in $\Bbb Z$. 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 $$(0,1)$$