# Increasing degree for Newton Interpolation

Given $$n+1$$ nodes $$x_0,...x_n$$ and a polynomial of degree $$n$$ called $$p_n$$ with $$p_i=f_i$$ for $$i=1,...,n$$. We looked into what happens when we increase the degree of $$p_n$$ to $$m:=n+1$$. It would be easy if $$p_m(x)=p_n(x)+g_m(x)$$ Now we know that that $$g_m(x_i)=p_m(x_i)-p_n(x_i) = f(x_i)-f(x_i)=0$$ for $$i=1,...,n$$. The notes says it follows that $$g_m(x)=a_m(x-x_0)\cdot...\cdot(x-x_n)$$ Now I want to understand how to get to this last conclusion. Obviously if we have $$n$$ roots $$x_0,...x_n$$ give for $$p_m$$ than these roots can be written as linearfactors. But this seems not very convincing and I wanted to ask if there is a way to show this using the Uniqueness of the interpolation polynomial?

$$g_m$$ is a polynomial of degree $$n+1$$ and by construction you know its $$n+1$$ roots at $$x_0,x_1,...,x_n$$. Then it is trivial that $$g_m$$ must be the product of the linear factors of the roots with a constant coefficient, as claimed.