Prove if $c$ is algebraic over $F$, so are $c + 1$ and $kc$ (where $k\in F$) This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise D1.

Prove that if $c$ is algebraic over $F$, so are $c + 1$ and $kc$ (where $k\in F$).

By definition, $c$ is algebraic over $F$ if it is the root of some nonzero polynomial $a(x)$ in $F[x]$.
Let $a(x)=a_0+a_1x+\cdots+a_nx^n$.
Consider $a(c+1) = a_0+a_1(c+1)+\cdots+a_n(c+1)^n = a(c) + b(c)$ for some nonzero polynomial $b(x)\in F[x]$ where $\operatorname{deg} b(x) < n$.
So $a(c) = a(c+1)-b(c) = 0$.  Suppose $b(x)=b_0+b_1x+\cdots + b_{n-1}x^{n-1}$. Let $b'(x) = b_0 + b_1(x-1)+\cdots +b_{n-1}(x-1)^{n-1}$ so that $b'(x+1)=b(x)$. 
$a(c+1)-b(c) = 0\implies a(c+1)-b'(c+1) = 0 = s(c+1)$ for some nonzero polynomial $s(x) = a(x) - b'(x)\in F[x]$. Note $s(x)$ is a nonzero polynomial for $\operatorname{deg} b'(x) < \operatorname{deg} a(x)$. Hence $c+1$ is algebraic over $F$.
If $k=1$, then $kc=c$ which is by assumption algebraic over $F$.
If $k=0$, then any nonzero polynomial $a(x)$ with a zero constant term can satisfy $a(kc) = 0$; so $kc=0$ is trivially algebraic over $F$.  Note as a nonzero polynomial that has a root, $a(x)$ must be a non-constant polynomial.
Suppose $k\ne 0$ and $k\ne 1$. $a(kc) = a_0+a_1(kc)+\cdots+a_n(kc)^n = a(c) + a_1(k-1)c +\cdots + a_n(k^n-1)c^n = a(c) + b(c)$ for
$b(x) = a_1(k-1)x +\cdots +a_n(k^n-1)x^n$.
If $b(x)$ is the zero polynomial, we are done, since $a(kc)=a(c)=0$.  
Suppose $b(x)$ is a nonzero polynomial in $F[x]$. So $a(kc) - b(c) = a(c) = 0$.  Let $\displaystyle b'(x)= a_1\frac{k-1}{k}x +\cdots + a_n\frac{k^n-1}{k^n}x^n$ so that $b'(kx) = b(x)$. 
$a(kc)-b(c) = 0\implies a(kc)-b'(kc) = 0 = s(kc)$ for some nonzero polynomial $s(x)=a(x)-b'(x)\in F[x]$. Note $s(x)$ is a nonzero polynomial for $\displaystyle\frac{k^i-1}{k^i} \ne 1$ for $0 < i \le n$.  Hence $kc$ is algebraic over $F$.
Does this look reasonable?
 A: Your solution is mixed up between polynomials $f(x)$ over $F$ where $x$ is a variable, and the value $f(c)$ of such a polynomial for some $c \in F$. For the first part you need to show that, if $a(x)$ is a polynomial function of $x$ with $c$ as a root, then the function $x \mapsto a(x - 1)$ is a polynomial function of $x$ that has $c + 1$ as a root. Similarly, for the second part. 
A: Let $c$ be algebraic over $F$, say $c$ is the root of $f(x) = a_0 + a_1x + \dots + a_nx^n$, that is, $f(c)= 0$.
To show $c + 1$ is a root, let $\beta = c + 1 \iff \beta -1 = c$ suggesting $f(x - 1)$ to be the desired polynomial. 
For $kc$, if $k = 0_F$, then $f(x) - a_0$ is the desired polynomial. If $k \neq 0_F$, then $\beta = kc \iff \beta/k = c$ suggesting $f(x/k)$ to be the desired polynomial.
A: Part one looks good. For part two, $b(x)$ may be the zero polynomial. For example, when $k=1$, $b(x)=0$. Even if $k\neq 1$, you still need to show why $b(x)$ is a non-zero polynomial. 
Edit:
If $k\neq 1$, $k^{n}-1$ can still be zero. If $a_{1},\ldots,a_{n-1}=0$, then $b(x)$ is still zero. 
