The basic answer is there is almost never cancellation. Following your example of $y=x^2$, you have the data points $(0,0),(1,1),(2,4)$ which it interpolates. Now if we pick one more point, say $(3,y)$ there will only be cancellation if $y=9$. For any other value, the interpolating polynomial will be of third degree. The same is true for $n$ points. There is a unique (at most) $n-2$ degree polynomial which interpolates the first $n-1$ points. There will only be cancellation if that polynomial passes exactly through the $n^{\text{th}}$ point.
Lagrange interpolation is nothing but a special case of CRT (Chinese Remainder Theorem). Namely, the special case where the ring is a ring of polynomials $\,K[x]\,$ over a field $\,K.$
Then we may apply CRT to the system $\ q(x) \equiv q(a_i) \pmod{x-a_i},\,$ because the $\,x-a_i\,$ are pairwise coprime (comaximal), since the $\,a_i\,$ are distinct, and the coefficient ring is field.
CRT yields a solution $\,q(x)\,$ that is unique modulo $\,g(x) = (x-a_1)\cdots (x-a_n).\,$ In particular there is a unique solution of degree $\,< \deg g.\,$ So one way to force your solutions to be unique is to reduce them mod $g$ to decrease the degree, and normalize the coefficients modulo $251,\,$ e.g. choose the least natural remainders mod $251$, i.e. those in the interval $[0,250],\,$ e.g. yours is
$\,{\rm mod}\ 251\!:\ {-51}\equiv 200,\,$ so $\,\color{#c00}{-51/2}\equiv 200/2\equiv \color{#c00}{100},\ $ and $\ \color{#0a0}{251\equiv 0}\ $ hence
$$\color{#c00}{-51/2}\,\ x^2 +\color{#0a0}{251}/2\,\ x + 100\, \equiv\, \color{#c00}{100}\, x^2 + 100$$
![Note I didn't use x_[0])](https://i.stack.imgur.com/MMeFP.gif)
Best Answer
Suppose we have $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$, the Lagrange interpolating polynomial is
$$p_4(x)=y_1 \frac{(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)}+y_2 \frac{(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_1)(x_2-x_3)(x_2-x_4)}+y_3 \frac{(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_1)(x_3-x_2)(x_3-x_4)}+y_4 \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_1)(x_4-x_2)(x_4-x_3)}.$$