Consider a function $y=f(x)$ that takes the values $$f(1)=0, \ f(2)=0.6931, \ f(4)=4.1589, \ f(3)=2.1972,$$ where $x_1=1, x_2=2, x_3=4, x_4=3.$ Show that the Lagrange polynomials $l_1,..,l_4$ defined by the above data satisfies $$l_1(x)+l_2(x)+l_3(x)+l_4(x)=1 \ \ \forall x. \tag{1}$$
Here we define the Lagrange polynomials, $l_j(x)$, by $$l_j(x)=\prod_{{k=1} \\ {k\neq j}}^{n+1}\frac{x-x_k}{x_j-x_k}.$$ I have computed the Lagrange polynomials $l_j(x)$ for $j=1,..,4$ as follows: \begin{align} l_1(x)&=-\frac{1}{6}(x-2)(x-3)(x-4) \\ l_2(x)&=\frac{1}{2} (x-1)(x-3)(x-4) \\ l_3(x)&=\frac{1}{6}(x-1)(x-2)(x-3) \\ l_4(x)&=-\frac{1}{2} (x-1)(x-2)(x-4). \end{align} Substituting the above equations into $(1)$ does appear to satisfy the required equation, but this method is computationally tedious. The answer provided in my textbook uses the fundamental theorem of algebra to explain this result, but I did not understand the explanation.
How can I go about answering this question?
The sum of the $l_j(x)$ is an at most cubic polynomial that has the value $1$ at all of $x=1,2,3,4$. The difference to $1$ is an at most cubic polynomial with $4$ roots, of which there is only one.