IS $f(x)= x^4+3x^2+2$ is irreducible over $\mathbb{Q}[X]$

Question

According to the definition that i have studied - A non constant polynomial f(x) in F[X] is said to be irreducible if f(x) cannot be expressed as a product of two polynomials of lower degree ,

now wat iam doing is - $f(x)= x^4+3x^2+2= (x^2+2) (x^2+1) = g(x).h(x)$ where $deg(g(x)) =2 < deg(f(x))$ and $deg(h(x))<deg((f(x))$ and both $g(x)$ and $h(x) \in \mathbb{Q}[X]$ , so according to the definition what i read this $f(x)$ is reducible over $\mathbb{Q}$ , Also $f(x)$ has no roots in $\mathbb{Q}$ , as $f(x)= x^4+3x^2+2= (x^2+2) (x^2+1) = (x+\sqrt2i) (x-\sqrt2i) (x+i) (x-i)$, my doubt is - if $f(x)$ is reducible over $\mathbb{Q}[x]$ then it must have roots in $\mathbb{Q}$ ????

Answer 1

So it's certainly not irreducible over $\mathbb{Q}$. Since $x^4+3x^2+2=(x^2+1)(x^2+2)$.

Answer 2

It certainly is reducible, because $f(x)=(x^2+2)(x^2+1)$.

In case you have intended an irreducible polynomial of degree $4$, compare with other questions on this site, e.g.,

$x^4-3x^2+4$ irreducible over over $\mathbb{Q}$

Answer 3

The problem of testing whether a polynomial is irreducible or not depends on the allowed field of coefficients. For example, the equation $x^2+3$ is irreducible over the field of rational numbers but is eminently reducible over irrationals by writing: $x^2-3=(x+\sqrt3)(x-\sqrt(3)$.

Similarly, $x^2+3$ is irreducible over bother rational and irrational field but is reducible over the complex filed as: $x^2+3=(x+i\sqrt3)(x-i\sqrt3)$.

Your factorisation above is obviously correct over the integer field but we could go further if the complex field is permitted.