I am given the following: Let $f(x)=x^4-a^2D$ such that $a$ and $D$ are nonzero, $a \in \mathbb{Q}, D \in \mathbb{Z}$.
In addition, I am given that $D$ is also square-free and $D \neq 1$
i) if $D \neq -1$ then $f(x)$ is irreducible over $\mathbb{Q}$
ii) if $D=-1$ then it is reducible over $\mathbb{Q}$ if and only if either $2a \in \mathbb{Q}^2$ or $-2a \in \mathbb{Q}^2$
I am not really sure how to do these. I have been using Eisenstein and the rational root test, but this is a more generic polynomial. Can someone demonstrate how to show these two things?
Here is a more general fact: if $F$ is a field whose characteristic is not $2$ and $\lambda\in F$, then $x^4-\lambda$ is reducible if and only if either $\lambda=r^2$ for some $r\in F$ or $-4\lambda=r^4$ for some $r\in F$. When $F$ has characteristic $2$, $x^4-\lambda$ is reducible if and only if $\lambda=r^2$ for some $r\in F$.
When you apply this result to the OP's example via $\lambda=a^2D$, the hypotheses imply $\lambda$ cannot be a square and $-4\lambda$ cannot be a fourth power if $D\ne -1$. If $D=-1$, then $-4\lambda=4a^2$, which is a fourth power precisely when $\pm 2a$ is a square.
To prove the initial assertion, note first that if $\lambda=r^2$, then $x^4-\lambda=(x^2-r)(x^2+r)$. If $-4\lambda=r^4$ and $F$ does not have characteristic $2$, then $x^4-\lambda=(x^2+rx+\frac12r^2)(x^2-rx+\frac12r^2)$.
For the converse, if $\lambda$ is not a square, then $x^4-\lambda$ has no roots, so any factorization of it must be equivalent to $(x^2+ax+b)(x^2-ax+d)$. Expanding this and equating it to $x^4-\lambda$ leads to the equation $a(d-b)=0$ (consider the coefficient of $x$), so either $a=0$ or $b=d$. When $a=0$, we have to have $d=-b$ and $b^2=\lambda$. When $a\ne 0$, we have $b=d$ and $a^2=2b$ (look at the coefficient of $x^2$). This immediately implies $F$ cannot have characteristic $2$. Since $bd=-\lambda$, we get $-4\lambda=a^4$.