A polynomial $P(x)$ of degree $n-1$ interpolates $n$ distinct points $(x_i, f_i)$, i.e $P(x_i) = f_i$, $i = 1,\dots, n$. Assuming that $P(x)$ is known, we wish to construct, without using the Lagrange polynomials $L_k(x)$, the polynomial of degree $n$, $Q(x)$, that interpolates the new point $(x_{n+1}, f_{n+1})$ in addition to the $n$ points aforementioned.
We will say: $Q(x) = P(x) + W(x)$, where $W(x)$ is a polynomial to determine.
How can I prove that the points $x_i$, $i=1,\dots,n$ are $n$ roots of $W$? Then, finding the degree of $W$, I want to be able to show that \begin{equation} W(x) = C \prod_{i=1}^n (x-x_i), \end{equation} where $C$ is a constant.
I am struggling to figure out how to show a few of these properties, so any guidance would be appreciated.
To show that every $x_i$ is a root to $W$, we can re-write the definition of $W$ as $$W(x) = Q(x) - P(x)$$ but $Q$ and $P$ are equal for all $x_i \ i = 1, 2, \ldots, n$ as they both interpolate the same $n$ points, so the claim follows.
For the degree of $W$, we use the fact that two polynomials are equal if and only if their coefficients corresponding to the same powers of $x$ are equal. As the left-hand side of $Q = P + W$ is a polynomial of degree $n$, the right hand-side must be, too. We know that $P$ is of degree $n-1$ so $W$ must have a $x^n$ term and no higher powers of $x$, proving the claim.
For the explicit formula for $W$, we can use the fundamental theorem of algebra and the fact that we know that $W$ is of degree $n$ and has $n$ real roots.