1. ## Irreduclible polynomials

How can I show that these polynomails are irreducible over rationals.

f(x) = x^4+x-1
g(x)= x^4+8x+12
h(x) = 1+x+x^2/2+x^3/6+x^4/24
j(x) = x^4-4x-1

2. Originally Posted by ZetaX
f(x) = x^4+x-1
Write $x^4 + x - 1 = (x^2 + ax + b)(x^2+bx+d)$ where $a,b,c,d\in \mathbb{Z}$ and show this is impossible by comparing coefficients.

You can try similar arguments for the others.

3. Hello, ZetaX!

I would assume you're expected to use the Rational Roots Theorem.

Given a polynomial $p(x)$, if $p(a) = 0$, then $(x-a)$ is a factor of $p(x).$

How can I show that these polynomials are irreducible over rationals?

$(1)\;\;f(x) \:=\: x^4+x-1$

$(2)\;\;g(x)\:=\: x^4+8x+12$

$(3)\;\;h(x) \:=\: 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\;\; {\color{blue}=\;\frac{1}{24}\left(x^4 + 4x^3 + 12x^2 + 24x + 24\right)}$

$(4)\;\;j(x) \:=\: x^4-4x-1$

All the polynomials have a leading coefficient of 1.
. . So the only possible rational roots are factors of the constant term.

For (1), the only choices are: . $\pm1\quad\hdots$ Try them in $f(x).$

For (2), the only choices are: . $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$

For (3), the only choices are: . $\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$

For (4), the only choices are: . $\pm1$

4. Originally Posted by Soroban
Hello, ZetaX!

I would assume you're expected to use the Rational Roots Theorem.

Given a polynomial $p(x)$, if $p(a) = 0$, then $(x-a)$ is a factor of $p(x).$

All the polynomials have a leading coefficient of 1.
. . So the only possible rational roots are factors of the constant term.

For (1), the only choices are: . $\pm1\quad\hdots$ Try them in $f(x).$

For (2), the only choices are: . $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$

For (3), the only choices are: . $\pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24$

For (4), the only choices are: . $\pm1$

Be careful if a fourth degree polynomial has no zeros does not mean it is irreducible.
Consider, $x^4+2x^2+1$, no rational zeros but it can be factored.