Proving is easy, but again I am not sure about the second part of the task
$ker(F) = $ {$f∈R_{n}[X]: F(f)=0$} = {$f = a_{0}$} , where $a_{0}$ can be any real number
So $ker(F) =$ {$a_{0}$, $a_{0}$∈$R$}, am I right?
And then $Im(F)$ using same "tactics" is gonna be equal to $R^{n-1}$? // (edit) It would dim(im(F)) not im(F) itself
hmm...
// $R_{n}[X]$ is set of all polynomials of degree equal or lower than $n$