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I have two questions. I want to prove $x^3-6x^2+12x+7$ is irreducible in rational numbers. My attempt is to use Gauss's lemma, that a primitive polynomial is irreducible in integers iff it is irreducible in rationals. And in integers, it is enough to prove there's no proper factorization of the polynomial. Next, I assume there is a linear factor, $(x-a)$. Then $a$ must divide $7$. But such an integer doesn't exist. Therefore, the polynomial has no linear factor, thus no quadratic factor. This implies $x^3-6x^2+12x+7$ has no proper factorization in integers, so it is irreducible in rationals.

Please tell me whether my answer is correct.

And for polynomial $x^4+8x^3+24x^2+35x-8$, I used a similar approach. First prove there's no linear factor, and then consider two quadratic factors. By equating the coefficients, I showed that there's no such integers. Then the polynomial has no proper factorization so irreducible.

I want to know does this approach generally work, and is there a technique to use Eisenstein's Criterion to show these two polynomials are irreducible.

Nonenicht
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  • This is not clear. Of course there are integers that divide $7$, we have $\pm 1$ and $\pm 7$. You need to eliminate those. – lulu Mar 28 '21 at 10:02
  • I mean by plugging in those 4 values, none of them is a root of the polynomial – Nonenicht Mar 28 '21 at 10:04
  • Nothing in what you wrote suggests that you tried that. – lulu Mar 28 '21 at 10:05
  • As to the Eisenstein question...well, it doesn't look good. Of course you can try to factor $\pmod p$ for various primes. But, for the cubic, the first prime for which it is irreducible is $p=13$ and, really, it's not that much easier to demonstrate that than it is to solve the original problem. – lulu Mar 28 '21 at 10:06
  • For the quartic, $p=11$ works, though again I doubt that this saves you much time. – lulu Mar 28 '21 at 10:08
  • I didn't show every single step above. I only want to know whether my approach works generally in these types of questions – Nonenicht Mar 28 '21 at 10:10
  • So, sure. For small degree polynomials it suffices to look at the various cases for factors. That method does not generalize terribly well to higher degree because the number of possible factorings grows too fast. The concept, however, would hold. – lulu Mar 28 '21 at 10:13
  • if I want to use Eisenstein's Criterion after shifting, is there a method to find that shift? For example, use x-a instead of x, how to find the a? – Nonenicht Mar 28 '21 at 10:16
  • Trial and error. As I say, working $\pmod p$ is a good technique which often works. And it works here, just not terribly easily. – lulu Mar 28 '21 at 10:19

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