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Let's define polynomial $P(x)$ as $ P(x)=x^3+2x^2-4x+2 $. My main question is to prove that it is irreducible and that its discriminant is $-140$, from the definition of polynomial discriminant and using cubic method of discriminant. Here is a picture:

$$D_3=a_1^2a_2^2-4a_0a_2^3-4a_1^3a_3+18a_0a_1a_2a_3-27a_0^2a_3^2$$

How can we show that it is irreducible? Thanks a lot.

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4  
Have you ever heard of Eisenstein's criterion? –  Jyrki Lahtonen Sep 22 '11 at 18:41
3  
Also the rational roots test will work here. –  Zarrax Sep 22 '11 at 19:27
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"Irreducible" is not an absolute; for example, since every cubic with real coefficients has a root in $\mathbb{R}$, $P(x)$ is not irreducible over $\mathbb{R}$; and the Fundamental Theorem of Algebra tells you it is not irreducible over $\mathbb{Q}$. You need to say "irreducible over something". In this case, it looks like you mean "over $\mathbb{Q}$". –  Arturo Magidin Sep 22 '11 at 19:48
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@Zarrax: A good point. I, too, realized that in this (=cubic) case Eisenstein's criterion really is just a fancy way of saying that the candidate rational roots won't work, because the values of the values of the polynomial at the candidate points are either odd or congruent to $2\pmod 4$ depending. –  Jyrki Lahtonen Sep 22 '11 at 20:05
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@Arturo: you must mean that the Fundamental Theorem of Algebra tells us it is not irreducible over $\mathbb{C}$. You were probably anticipating the next two sentences. –  robjohn Sep 22 '11 at 21:19

3 Answers 3

up vote 4 down vote accepted

HINT $\ $ If reducible it would have an integer root, hence a root mod $3$. But it has no roots mod $3$.

In fact $\rm\: f(x) =\: x^3 + a\ x^2 + b\ x + c\ $ is irreducible over $\rm\:\mathbb Q\ $ if $\rm\ c\:\not\equiv\: 0\ $ and $\rm\:a+c\ \not\equiv\: \pm\: (1+b)\pmod{3}$

The first condition implies that $\rm\:f(0)\:\not\equiv\:0\:,\:$ and the second that $\rm\:f(\pm1)\:\not\equiv\:0\pmod 3\:.$

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Eisenstein's Criterion says that since $2$ divides all of the coefficients but the lead coefficient and $2^2$ does not divide the constant coefficient, $P$ is irreducible over $\mathbb{Q}$.

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As @zarrax pointed out, the rational roots test suggests that if $p/q$ is a rational root of $P(x)=x^3+2x^2-4x+2$ then $p|2$ and $q|1$ which means that $p=\pm 1, \pm 2$ and $q=\pm 1$. Now since all $p/q=-1,1,-2,2$ are not roots of $p(x)$, it follows that the polynomial $p(x)$ is irreducible over $\mathbb Q$.

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