Solutions

We have

$$d'(x)=f'(x)-f'(a),$$

so $d'(a)=0$.

Such that $n-1-k$ is even. Hence, $f(x)=0$ for precisely $k$ numbers $x$. Moreover, by Problem 10-22 a), there exists a polynomial function $g$ of degree $n$ such that $g'=f$. In this case, it follows that $n-k$ is odd since $n-1-k$ is even.