Proving fundamental properties of polynomials My question is from Apostol's Vol. 1 One-variable calculus with introduction to linear algebra textbook.
Page 57. Exercise 9. This exercise develops some fundamental properties of polynomials. Let $f(x)=\sum_{k=0}^nc_kx^k$ be a polynomial of degree $n$. Prove each of the following:
a) If $n\ge1$ and $f(0)=0$, then $f(x)=xg(x)$, where $g$ is a polynomial of degree $n-1$.
b) For each real $a$, the function $p$ given by $p(x)=f(x+a)$ is a polynomial of degree n.
c) If $n\ge1$ and $f(a)=0$ for some real $a$, then $f(x)=(x-a)h(x)$, where $h$ is a polynomial of degree $n-1$. [Hint: Consider $p(x)=f(x+a)$.]
d) If $f(x)=0$ for $n+1$ distinct real values of $x$, then every coefficient $c_k$ is zero and $f(x)=0$ for all real $x$.
e) Let $g(x)=\sum_{k=0}^mb_kx^k$ be a polynomial of degree $m$, where $m\ge n$. If $g(x)=f(x)$ for $m+1$ distinct real values of $x$, then $m=n, b_k=c_k$ for each $k$, and $g(x)=f(x)$ for all real $x$.
The attempt at a solution: I proved part a), I think I proved b) as well, I have trouble proving c) and e). Please comment if you need information about how I proved first 2 parts of the problem. Any help would be appreciated, thank you.
Proof of a): We have $f(x)=\sum_{k=0}^nc_kx^n$ which is true for $n\ge1.$
Since $$f(0)=c_0x^0+c_1x+c_2x^2+\cdots+c_nx^n=0,$$since $0^0=1$, we have $$c_0*1=0,$$$$c_0=0.$$ We also have $$g(x)=c_0x^{-1}+c_1x^0+c_2x^1+\cdots+c_nx^{n-1},$$and$$f(x)=xg(x),$$$$\sum_{x=0}^nc_kx^k=x\sum_{k=0}^nc_kx^{k-1},$$$$c_0x^0+c_1x+c_2x^2+\cdots+c_nx^n=x(c_0x^{-1}+c_1x^0+c_2x^1+\cdots+c_nx^{n-1}),$$since $c_0=0$, we can write $$c_1x+c_2x+\cdots+c_nx^n=c_1x+c_2x+\cdots+c_nx^n.$$
proof of b): We know that $$p(x)=f(x+a)=\sum_{x=0}^nc_k(x+a)^k,$$$$p(x)=c_1(x+a)+c_2(x+a)^2+\cdots+c_n(x+a)^n,$$which is polynomial of degree $n$.
Sorry if my proofs aren't very formal, I haven't had proper training in proofs as you can see, but I'm trying.
 A: (c) With $p(x)=f(x+a)$, if $f(a)=0$, then $p(0)=0$. By part (a), we can write $p(x)=xg(x)$ for some $x$. But then:
$$
f(x)=f(x-a+a)=p(x-a)=(x-a)g(x-a)=(x-a)h(x)
$$
where $h(x)\equiv g(x-a)$.
(e) Let $\phi(x)=g(x)-f(x)$ be a polynomial of degree $m$. Then use (d) to infer that $\phi(x)=0$ for all $x$ and the coefficients of $\phi$ are $0$. The former gives $f(x)=g(x)$ $\forall x$ whereas the latter gives $m=n$ and $b_k=c_k$ for each $k$.

Edit: you also asked for (c) and (e) but for completeness, here is (d):
(d) Let the distinct roots be $a_1,\ldots, a_{n+1}$. Using (c) $n$ times, we get $f(x)=(x-a_1)\cdots (x-a_n)c$ where $c$ is a constant. But $c$ has to be $0$ because $f(a_{n+1})=0$ and $a_{n+1}$ is distinct from $a_1,\ldots,a_n$. This in turn implies that $f(x)=0$ everywhere.
To show $c_0=\ldots=c_n=0$, we can proceed as follows. Let $f_0=f$. By the previous paragraph, we have $f_0(0)=0$ so $c_0=0$ and we can write $f_0(x)=xf_1(x)$ where $f_1$ is a polynomial of degree $n-1$ with coefficient $c_1$ associated with $x^0$. Plug some $n$ distinct nonzero numbers into $f_0$, we can infer from $f_0=0$ everywhere that $f_1$ has $n$ distinct roots. We can therefore repeat the argument thus far to infer that $c_1=0$. Continuing like this gives $c_0=\ldots=c_n=0$.
