Deriving the Langrangian interpolation polynomials in Cook-Toom convolutions
If you represent a second-order polynomial $s(x)$ with Lagrange polynomials $L_i(x)$ and interpolation points $\beta_i$, $i=0,1,2$, such that

$$s(x)=s(\beta_0)L_0(x)+s(\beta_1)L_1(x)+s(\beta_2)L_2(x)\tag{1}$$

then for the equality in $(1)$ to be satisfied, the polynomials $L_i(x)$ must have zeros at $x=\beta_j$, $j\neq i$, and they must equal $1$ at $x=\beta_i$. This completely determines the polynomials $L_i(x)$.

E.g., the polynomial $L_0(x)$ must have two zeros at $\beta_1=1$ and $\beta_2=-1$, respectively, and it must satisfy $L_0(\beta_0)=L_0(0)=1$. The latter condition just determines the scaling. This results in

$$\begin{align}L_0(x)&amp;=\frac{(x-\beta_1)(x-\beta_2)}{(\beta_0-\beta_1)(\beta_0-\beta_2)}\\&amp;=-(x-1)(x+1)\\&amp;=1-x^2\end{align}$$

The other polynomials can be derived in the same way.

Note that the polynomials $L_1(x)$ and $L_2(x)$ have a scaling factor of $\frac12$, which is later in the derivation absorbed in other constants for reasons of computational efficiency.