Is $4x^2−1$ irreducible in $\mathbb{Z}[x]$?

Question

Is $4x^2−1$ irreducible in $\mathbb{Z}[x]$ .

I think yes as the roots of the polynomial are not integers so it is irreducible in $\mathbb{Z}$, but it is reducible in $\mathbb{Q}[x]$ as we could factor out the polynomial as a product of two linear factors and the roots are rational so it is reducible in $\mathbb{Q}[x]$?

Next I encounter this theorem If $f(x)$ is irreducible in $D[x]$ ($D$ is an Integral domain) then it is also irreducible in $F[x]$,where $F$ is the field of Quotients of $D$

now comparing this above as $4x^2−1$ is irreducible in $\mathbb{Z}[x]$ it must be irreducible in $\mathbb{Q}[x]$ , but we saw above that it is reducible in $\mathbb{Q}[x]$?

Any mistake here?

Answer 1

No, $4x^2- 1$ is reducible in $\mathbb{Z}[x]$ as we can write the polynomial as linear factors as $(2x + 1)(2x - 1)$ which shows reducibility in $\mathbb{Z}[x]$.

But we cannot factor the above polynomial as $4(x - \frac{1}{2})\cdot(x + \frac{1}{2})$ as factoring that requires the inverse of the $x^2$ term coefficient which doesnot exist in $\mathbb{Z} $ as $\mathbb{Z}$ is not a field.

So the above polynomial is both reducible in $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$.

Answer 2

The answer is no. Infact $$4x^2-1=(2x-1)(2x+1)$$

and both factors belong to $\Bbb Z[x]$.