Irreducibility of a Polynomial in Q[x]

Question

let f(x)=2x^2-8 is a polynomial in Q[x].Although we can chech for reducibility by checking the zero of the polynomial in Q(because it is a field) and we see that x=2 is a zero of this. but if we write f(x)=g(x)h(x) =2(x^2-4) then we see that 2 is unit in Q does that mean f(x) is irreducible. And the definition is if non zero non unit a is irreducible if a=bc then either of them is unit. what am I missing here? please explain.

Answer 1

You have to think and speak with quantifiers, otherwise you won't understand what you are reading (and, which is arguably worse, you might say three different things every three times you speak). A polynomial $p$ is irreducible if and only if it is not invertible and, for all $f$ and $g$ such that $fg=p$, either $f$ is invertible or $g$ is invertible.

Therefore existence of an invertible element $f$ and a polynomial $g$ such that $p=fg$ is completely inconsequential.

Answer 2

but if we write f(x)=g(x)h(x) =2(x^2-4) then we see that 2 is unit in Q does that mean f(x) is irreducible

The definition says that $f$ is irreducible if no decomposition $f=gh$ into non-zero non-units is possible. To phrase it in terms of reducibility, a polynomial $f$ is reducible if there exists a decomposition $f=gh$ such that $g$ and $h$ are non-zero non-units.

You have chosen $g(x)=2$ and $h(x)=(x^2-4)$, so you have given one decomposition into a product of a unit and a non-unit. But that does not prove that there are no other choices for $g$ and $h$ where both are non-zero non-units, for example $g(x)=2(x-2)$ and $h(x)=(x+2)$.

(Also, think of the following: if writing a polynomial $f$ as a product of a unit $g$ and a non-unit $h$ would prove its irreducibility, then all polynomials would be irreducible since $f=1\cdot f$, so we wouldn't distinguish between the concepts of irreducible and reducible polynomials in the first place).