I would like to see if I'm right about these polynomials I tried to prove are irreducible:

1) For the first polynomial I used that if a polynomial is irrational over $\mathbb{Z}_p$ for $p$ prime, and $p$ does not divide the $x^n$ coefficient of this polynomial (suppose degree $n$), then it is irreducible over $\mathbb{Q}$.

$$p(x) = x⁴+2x³+2x²+2x+2$$

I choose $\mathbb{Z}_2$. Well, over $\mathbb{Z}_2$ we get:

$$p(x) = x⁴$$

the only possible root for this polynomial is $x=0$, because $1⁴=1$. Can I say this polynomial is irreducible because its only root is $0$ and I would break it into $(x-0)(x-0)(x-0)(x-0)$ which is just $x⁴$?

2) For the second polynomial I used the Eisenstein's criteria: if $p$ is a prime such that $p²$ does not divide $a_0$, but not $a_n$ and $p$ divides the other coefficients, then $p(x)$ is irreducible over $\mathbb{Q}$:

$$p(x) = x^7-31$$

it's clear that $p=31$ will divide $a_0$ but $p²$ will not, and also $p$ does not divide $a_n$. In fact, every polynomial in the form $x^n-p=0$ will be irreducible over $\mathbb{Q}$ by this criterion

3) $$p(x) = x^6+15$$

We have that $p=3$ will divide $a_0=15$ but $p^2=9$ will not. Also, $p=3$ will not didivde $a_n=1$, so $p$ is irreducible over $\mathbb{Q}$.

4) $$p(x) = x^3+6x^2+5x+25$$

I cannot use the criterion here because $p$ must divide $25$ so $p=5$ but $p$ will not divide $a_2 = 6$. So I'll try to reduce it $mod p$. Taking it $mod 5$ I get:

$$p(x) = x^3+x^2$$

well, it didn't work. So I'll try $mod 3$, we get:

$$p(x) = x^3+2x+1$$

which is reducible...

So let's try $mod 2$:

$$p(x) = x^3+x+1$$

which is still reducible :c

Any ideas on this one?

5) $$p(x) = x^4+8x^3+x^2+2x+5$$

if we reduce mod $2$ we get:


which I suspect is irreducible. It's always positive, so it has no roots, so it can't be factored into at least one $1$ degree factor. The only possibility would be to factor it in two irreducible $2$ degree polynomials, which I'll try later to prove. Wolfram Alpha didn't give its factors so I think it's a signal.


$$p(x) = x^4+10x^3+20x^2+30x+22$$

reduction mod $2$ would be a good idea, but we have that $2$ divides $a_0$ but $2^2$ will not divide it. $2$ divides every other coefficients except $a_n=1$ so this polynomial is irreducible

  • $\begingroup$ 1) No, it is reducible over $\mathbb{F}_2$, which does not help. $\endgroup$ Commented Jan 9, 2016 at 19:43
  • $\begingroup$ @DietrichBurde can I say that $2$ divides $a_0=2$ but $2^2$ not, and that $2$ divides every other except $a_n$, so it is irreducible? If you had to reduce it, which $p$ would you take? $\endgroup$ Commented Jan 9, 2016 at 19:46
  • $\begingroup$ Yes. In 4), why do you think that $x^3+x+1$ is reducible modulo $2$ ? Same question for $x^3+2x+1$ modulo $3$ ? $\endgroup$ Commented Jan 9, 2016 at 19:48
  • $\begingroup$ @DietrichBurde I've put them in Wolfram Alpha and it had roots, but I forgot I was in $\mathbb{Z}_2$ and $\mathbb{Z}_3$, I should try its values and see if I have roots, other else this polynomial can't be reduced because it must factor in: one degree and two degree $\endgroup$ Commented Jan 9, 2016 at 19:56

2 Answers 2


(1) is screaming for you to use eisenstein's criterion since everything has a factor of $2$ except the first term.

(4) is irreducible mod $2$, if you plug in either $0$ or $1$ we get a value not congruent to $0$ mod $2$.

For (5) yes you should check to see whether it factors as 2 quadratic polynomials, and whether it has a rational root. Or just find a $p$ such that it is irreducible in $\mathbb{Z}/(p)$ and check that it does not factor as the product of two quadratic polynomials.

  • $\begingroup$ (5) is not irreducible mod 2. $\endgroup$
    – Bernard
    Commented Jan 9, 2016 at 21:14
  • $\begingroup$ Whoops thanks for noticing. Will edit $\endgroup$ Commented Jan 9, 2016 at 21:15

(1) has no rational root, since it would have to be a negative divisor of the constant term, and neither $-1$ nor $-2$ are roots. Hence if it were reducible, it would factor as the product of two quadratic polynomials, and we would have such a factorisation modulo $3$.

However, a factorisation over $\mathbf F_3$ of $\;p(x)=x^4-x^3-x^2-x-1$ as $$x^4-x^3-x^2-x-1=(x^2+ax+b)(x^2+a'x+b'$$ would lead to the system of equations $$a+a'=-1,\quad aa'+b+b'=-1,\quad ab'+ba'=-1,\quad bb'=-1$$ The last equation implies one of $b, b'$ is equal to $1$, the other to $-1$. Say $b=1,\enspace b'=-1$. The last but one equation becomes $a'-a=-1$, which is incompatible with the first equation. Thus $p(x)$ cannot split as the product of two quadratics over $\mathbf F_3$, and **a fortiori*, over $\mathbf Q$.

(4) is irreducible over $\mathbf Q$, since it has no rational root: such roots would be one of $-1, -5, -25$, and none of these values are roots.

For (5), same method as for (1): it ha no rational root, hence would split as the product of two quadratics, which is tested impossible reducing modulo $3$.

(6): reducing modulo $5$, we obtain the polynomial becomes $x^4+2 $ in $\mathbf F_5$, which has no root. The factorisation as a product of qudratics yields the system of aquations: $$ a+a'=0,\quad aa'+b+b'=0,\quad ab'+ba'=0, \quad bb'= 2. $$ From first and third relations, we deduce $a'=-a$ and $\;a(b'-b)=0$, hence

  • either $a=a'=0$, which implies (2nd relation) $b'=-b$, hence (4th relation) $b^2=-2$. However, $-2$ is not a square in $\mathbf F_5$.
  • or $b=b'$, whence $b^2=2$. But $2$ is not a square in $\mathbf F_5$.

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