Problem :
If $f(x)$ is a polynomial of degree $n$ and if $f(k) = \frac{k}{k+1}$ where $k =0,1,2,\ldots,n$, find $f(x)$.
Can we go like this : Let the polynomial be $a(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)......(x-\alpha_n)$ where $\alpha_1, \alpha_2....\alpha_n$ are the roots of the $f(x)$ and $a$ is the leading coefficient.
Please guide how to proceed in such question.
Hint: Consider the polynomial $g(x) = (x+1) f(x) - x$. What is the degree of $g(x)$? What are the zeros of $g(x)$? What is the value of $g(-1)$?
More details added.
The degree of $g(x)$ is equal to the degree of $f(x)$ plus 1 (since we multiplied by $x+1$), which is equal to $n+1$.
The zeros of $g(x)$ are (at least) $0, 1, 2, \ldots n$, so by the Remainder-Factor Theorem,
$$g(x) = A(x) \prod_{i=0}^n (x-i)$$
Since $g$ has degree $n+1$, it follows that $A(x)$ is a constant polynomial, which we will denote by $A$.
The value of $g(-1)$ is $g(-1) = (-1 + 1) f(-1) - (-1) = 1 $. Hence
$$1 = g(-1) = A \prod_{i=0}^n (-1-i)$$
This allows us to determine that $A = \frac{1}{(n+1)!}$.
Hint: $\frac{ g(x) + x) } { x+1}$ is indeed a polynomial, and is our $f(x)$.