Show that polynomial is irreducible over $\mathbb{Q}$

Question

How I can prove that polynomial $f(x)$, where$$f(x) = x^4 + 3x^3 + 3x^2 - 5$$ is irreducible over $\mathbb{Q}$? Thank you

Answer 1

$x^4 + 3x^3 + 3x^2 - 5$ is reducible if it has a rationnal or if it can be written in the form : $(x^2+ax+b)(x^2+cx+d)$.

Rationnal roots : Since the polynomyal is unitary ans has integer coefficients, the rationnal roots are integers. Furthemore such roots must divide $-5$. So let's test if $\exists x \in \{-5,-1,1,5\}$ such that $f(x)=0$. After calculations, we can conclude that $f$ does not have a rationnal root.

$f(x)=(x^2+ax+b)(x^2+cx+d)$ : It leads to $c+a=3,b+d+ac=3,ad+bc=0,bd=-5$. You can try to solve this system (it has no solution in $\mathbb{Q}$).

So the polynomial is irreducible over $\mathbb{Q}$.

Answer 2

There are several ways to show that $f(x)$ is irreducible. Since it has no rational root (rational root test), it could only be reducible as a product of two monic quadratic polynomials, $$ f(x)=(x^2+ax+b)(x^2+cx+d), $$ where we may assume that $a,b,c,d$ are integers (Gauss lemma). Comparing coefficients gives four Diophantine equations, which are easily shown to be inconsistent. They are given by $$ a+c-3=0,\; ac + b + d - 3=0,\; ad + bc=0,\; bd + 5=0. $$

Answer 3

$x^4 + 3x^3 + 3x^2 - 5$ is irreducible mod $7$.

To establish this, it suffices to test a finite number of linear and quadratic factors, as listed by WA.