I know that in $\Bbb Z[x]/\langle x^2 - 1\rangle$ the trivial idempotents are $0,1$ and the other idempotents are those elements in $\Bbb Z[x]$ that have the remainder $0$ or $1$ when divided by $x^2 - 1$. I end up with $x + 1$ and $x$ being the other idempotents.
I have no idea about $\Bbb Q[x]$, since now you can have fractional coefficients for the polynomials.
Any help will be greatly appreciated.
On
A “direct” proof. Not that Martin's is bad, to the contrary it's very nice.
The ring $\mathbb{Q}[x]/\langle x^2-1\rangle$ can be described as the set of expressions $$ a+br $$ with $a,b\in\mathbb{Q}$, where $r\notin\mathbb{Q}$ satisfies $r^2=1$. Moreover $a+br=a'+b'r$ if and only if $a=a'$ and $b=b'$. Then $$ (a+br)^2=a^2+2abr+b^2r^2=(a^2+b^2)+2abr $$ and the element is idempotent if and only if $$ \begin{cases} a^2+b^2=a\\ 2ab=b \end{cases} $$ If $b=0$, we get $a^2=a$, that is, $a=0$ or $a=1$. If $b\ne0$ we get $a=1/2$ and $b^2=1/4$, that is $b=1/2$ or $b=-1/2$.
By the Chinese Remainder Theorem, $$\mathbb{Q}[x]/(x^2-1) = \mathbb{Q}[x]/((x+1)(x-1)) \to \mathbb{Q}[x]/(x+1) \times \mathbb{Q}[x]/(x-1) \cong \mathbb{Q} \times \mathbb{Q}$$ is an isomorphism. The proof is constructive and therefore includes the construction of the isomorphism as well as its inverse. Have a look at the proof. Explicitly, the isomorphism maps $[f]$ to $(f(-1),f(1))$ and the inverse maps $(1,0)$ to (the coset of) $\frac{1-x}{2}$ and $(0,1)$ to $\frac{x+1}{2}$, hence $(a,b) \in \mathbb{Q}^2$ to $\frac{a(1-x)+b(x+1)}{2}$. The product $\mathbb{Q} \times \mathbb{Q}$ has $4$ idempotents, namely $(a,b)$ with $a,b \in \{0,1\}$. Now we can just write down the corresponding idempotents in $\mathbb{Q}[x]/(x^2-1)$. These are $0,1,\frac{1-x}{2},\frac{x+1}{2}$.