I tried for few primes but they do not satisfy Eisenstein criterion.Also is there any approach other than brute force with the help of which we can find that prime.
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Constant term is -1, and no prime divides -1. You'll have to use something like x=t+a for some appropriate choice of a, to get Eisenstein's criterion to work. – Locally unskillful Apr 04 '19 at 18:29
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You can find further examples in the linked dupe. The method I describe there yields a simple algorithm to find such a prime (or prove that it does not exist). – Bill Dubuque Apr 04 '19 at 18:54
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1Note also that none of $-1$ or $1$ is a root, so you can use rational root theorem to get the irreducibility in this case. – Sil Apr 04 '19 at 21:41
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You can use the reducite criterium.
I.e., consider the polynomial over the field $\mathbb{F}_2$.I.e., we get $X^3 + X^2 +1$. If this reduced polynomial is irreducible over $\mathbb{F}_2$, it is also irreducible over $\mathbb{Q}$ (this theorem is known as the reducite criterium).
This polynomial is irreducible over $\mathbb{F}_2$ (it has no roots, this is easily seen).
Hence, by the reducite criterium it follows that $X^3 + X^2 - 2X -1$ is irreducible in $\mathbb{Q}[X]$.
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We don't need to pull the Eisenstein shift out of a hat like magic. I explain how to find it below.
Hint $\, \bmod\color{#c00}7\!:\,\ f(x) \equiv (x\!-\!2)^{3} \ $ is a prime power.
So Eisenstein works on $\,g(x) = f(x\!+\!2) \equiv x^3\ $ by $\,g(0) = f(2)\equiv 7\not\equiv 0\pmod{\!49}$
Remark $\ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $\,p,\,$ powers of a prime, viz. $\,\equiv x^n,\,$ and products of primes always factor uniquely. The same works for its shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $\,p\,$ that can yield such powers are those dividing the discriminant (here by Alpha = $\,\color{#c00}7^2).\,$ Indeed, if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$ hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).
Bill Dubuque
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Apply Eisenstein's criterion to $(x+2)^3+(x+2)^2-2(x+2)-1$, with $p=7$.
Mark Fischler
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José Carlos Santos
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Also is there any approach other than brute force with the help of which we can find that prime and $t$ in $(x+t)$? – Apr 04 '19 at 18:40
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1@Mahesh I explained that in my answer. There is even a link to the Alpha computation. – Bill Dubuque Apr 04 '19 at 18:40
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How did you know to substitute $x$ with $x + 2$ and to use $p = 7$?? – Junglemath Jun 24 '20 at 13:08
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@Junglemath I did not know. I tried $x+1$, $x-1$, and so on. – José Carlos Santos Jun 24 '20 at 14:59