Similar problem to Taylor's theorem proof 

43. Let $a_1,\dots,a_{n+1}$ be arbitrary points in $[a,b]$ and let
  $$Q(x)=\prod_{i=1}^{n+1}(x-x_i).$$
  Suppose that $f$ is $(n+1)$-times differentiable and that $P$ is a polynomial function of degree $\le n$ such that $P(x_i)=f(x_i)$ for $i=1,\dots,n+1$. Show that for each $x$ in $[a,b]$  there is a number $c$ in $(a,b)$ such that
  $$f(x)-P(x)= Q(x) \cdot \frac{f(n+1)(c)}{(n+1)!}.$$
  Hint: consider the function
  $$F(t)=Q(x)[f(t)-P(t)]-Q(t)[f(x)-P(x)].$$
  Show that $F$ is zero at $n+2$ different points in $[a,b]$, and use Problem 42.

Problem 42 was: 

"Suppose that $f$ is $n$ times differentiable and that $f(x) = 0$ for $n+1$ different $x$. Prove that $f^{(n)}(x) = 0$ for some $x$. "

I will start:
$$F(t) = Q(x)[f(t) - P(t)] - Q(t)[f(x) - P(x)]$$
$$F(a) = Q(x)[f(a) - P(a)] - Q(a)[f(x) - P(x)]$$
$$Q(a)  = \prod_{k=1}^{n+1} x- x_k = \prod_{k=1}^{n+1} a- a_k = (a- a_1)(a - a_2).... (a - a_{n+1})$$
But I don't see any options? Can someone guide me? Thanks!
PLEASE DO NOT GIVE A FULL ANSWER
 A: $\textbf{Hint:}$


*

*Check that $F(t)=0$ for $x_1,x_2,\ldots,x_{n+1},x$.

*Use Rolle's theorem $n+1$-times.

A: This is a jigsaw puzzle, and there are only finitely many ways to fit the pieces together.  You are given the statement:

Problem 42 Suppose that $f$ is $n$ times differentiable and that $f(x)=0$ for $n+1$ different values of $x$.  Then $f^{(n)}(x) = 0$ for some $x$.

Let's rewrite that slightly:

Suppose that $f$ is $n$ times differentiable and
  $$
f(x_1) = f(x_2) = \cdots = f(x_n) = f(x_{n+1}) = 0
$$
  for $n+1$ distinct points $x_1,x_2,\dots,x_{n+1}$.  Then there exists a point $c$ such that $f^{(n+1)}(c) = 0$.

You are given a function $F(t) = Q(x) [f(t)-P(t)] - Q(t) [f(x) - P(x)]$ and directed to show that there are $n+2$ distinct values of $t$ for which $f(t) = 0$.  Then by Problem 42 there will be a point $c$ such that $F^{(n+1)}(c) = 0$.  If you are lucky $F^{(n+1)}(c)=0$ will imply that 
$$
f(x) - P(x) = Q(x) \cdot\frac{f^{(n+1)}(c)}{(n+1)!}
$$
The only question is which $n+2$ values of $t$ will yield $F(t) = 0$?  There are $n+1$ points $x_1,x_2,\dots,x_{n+1}$ given.  Do they satisfy this equation?  Remember that you know something about $Q(x_i)$ and $P(x_i)$.  Can you find one more value of $t$?
