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I need to evaluate for which $a \in \mathbb{Z}$ the polynomial $3x^3+20ax^2+50a^2x+60$ is irreducible respectively over $\mathbb{Q}, \mathbb{C}$ and $\mathbb{R}$. I think it is a tricky question but I need not to be wrong. For the fundamental theorem of algebra in $\mathbb{C}$ there should be $n$ roots, so it's always reducible. Every polynomial splits into at worst quadratic factors in $\mathbb{R}$.

Given the fact that

$3x^3+a[20x^2+50ax]+60 $

if there was not the 60 I could find some $a$ such that $3x^3 = -20ax^2$ and get a polynomial of n=2 such that $\Delta \lt 0$ but it's not the case.

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    $\begingroup$ If it's reducible (0ver the rationals), it has a rational root. If it has a rational root, the numerator of that root divides $60$, and the denominator divides $3$. So we've reduced it to a finite problem. $\endgroup$ Oct 5, 2020 at 12:06
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    $\begingroup$ So, can you solve it now? $\endgroup$ Oct 7, 2020 at 10:25

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I have finally got it (I think). Every cubic polynomial is reducible in both $\mathbb{R}$ and $\mathbb{C}$.

With regards to rationals, either you use the rational test as suggested in the comment for all possible candidates

{-60, -30, -20, -15, -12, -10, -20/3, -6, -5, -4, -10/3, -3, -2, -5/3, -4/3, -1, -2/3, -1/3, 1/3, 2/3, 1, 4/3, 5/3, 2, 3, 10/3, 4, 5, 6, 20/3, 10, 12, 15, 20, 30, 60}

(if you find 3 roots you should have done since it's cubic)

or you notice that you can apply Eisenstein's test with $a_0 = 60$, $a_n = 3$, and $p=5$ to find out it is irreducible.

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  • $\begingroup$ Good! Eisenstein definitely better idea than testing all candidate rationals for this problem. $\endgroup$ Oct 8, 2020 at 22:17

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