Prove $X^4-5X^3+3X-2$ is irreducible over the rationals.

Question

How can we show that $P=x^4-5x^3+3x-2$ is irreducible over the rationals. I tried to use Eisenstein criterion and many shifts $P(x+a)$ but did not work. I tried to show it is irreducible modulo $2$ or $3$ but it did not work.

I tried to use the rational root test to prove that $P$ does not have any rational root. Also $P(-1)=1$, $P(0)=-2$, $P(1)=-3$ and $P(5)=13$. So we know that there are two irrational roots $\alpha\in (-1,0)$ and $\beta \in (1,5)$ so $\beta \not = -\alpha$, the other two roots are conjugate complex but I don't know how to prove it, is it enough to say that $P$ has two irrational non opposite roots and two conjugate complex roots to say it is irreducible over $\mathbb Q$?

Answer 1

By Gauss' lemma, irreducibility over $\mathbb{Q}$ is equivalent to irreducibility over $\mathbb{Z}$ in this case. Since there are no rational roots, the only way $f$ can be reducible is being a product of two monic polynomials $g, h\in\mathbb{Z}[x]$ of degree $2$. But now reduce the equation $f=gh$ modulo $2$. You get that the polynomial $\overline{f}=x^4+x^3+x\in\mathbb{F_2}[x]$ is a product of two monic polynomials of degree $2$ in $\mathbb{F_2}[x]$. But that is a contradiction, because:

$\overline{f}=x(x^3+x^2+1)$

is the factorization of $\overline{f}$ into irreducibles in $\mathbb{F_2}[x]$, and such factorization is unique. (if it was a product of two polynomials of degree $2$, it wouldn't have an irreducible factor of degree $3$)

Answer 2

Using Gauss' lemma, if a factorization exists, it has integer coefficients. The rational root theorem easily shows there is no integer root, so no factorization $(x-a)(x^3+bx^2+cx+d)$. There is still the possibility that $P=(x^2+ax+b)(x^2+cx+d)$, with integer coefficients. Then we have the equations, after expanding the product:

• $a+c=-5$
• $b+ac+d=0$
• $ad+bc=3$
• $bd=-2$

Since coefficients are integers, we can assume $b=\pm1$, then $d=\mp2$.

If $b=1$, the third equation becomes $-2a+c=3$, and together with the first we must have $3c=-7$, i.e. $c$ is not an integer, contradiction.

If $b=-1$, the third equation becomes $2a-c=3$, and together with the first, we get $3a=-2$, and $a$ is not an integer, again a contradiction.

Therefore there is no such factorization, and $P$ is irreducible.

Answer 3

By Rouché's theorem $P$ has exactly one zero outside the complex unit disc (see below). If $P$ was reducible into two monic polynomials with integral coefficients (justified by Gauss's lemma), one of the factors, say $Q(x)$, would have all roots $|\alpha_i|<1$. So we would get $|Q(0)|=|\prod \alpha_i|<1$, but that is clearly impossible since $Q(0)$ must be an integer that divides $2$. So $P$ must be irreducible over rationals.

Note to Rouché's theorem: For $|z|=1$ we have $|3z-2|<|z-5|=|z^4-5z^3|$ (the inequality defines disc with center $(\frac{1}{8},0)$ and radius $\frac{13}{8}$ which strictly contains a unit disc). Therefore $x^4-5x^3=x^3(x-5)$ and $P(x)$ have the same number (three) of roots inside the unit disc.