Assume that $u$ is algebraic over the field $F$, and that $a\in F$. Show that $u + a$ is algebraic over $F$, find its irreducible polynomial over $F$ and show that deg($u + a,F$) = deg($u,F$).
I made some observations:
$u$ is algebraic over $F \Rightarrow$ $f(u)=0$ for some nonzero $f(x)\in F[x]$. Let $n$=deg($u,F$). Then $u^n+a_{n-1}u^{n-1}+...+a_0=0$.
Now I tried to manipulate with $(u+a)^m+b_{m-1}(u+a)^{m-1}+...+b_0$. Since $F$ is a filed, we can use binomial theorem. Goal is to show $n=m$ and to find structure of this polynomial. I'm not sure if I can approach with some kind of $g(u)=f(u+a)$. I'm a little bit confused about how to proceed with all of this.
Replacing $x$ with $x-a$ in $f$ yields the polynomial $g(x)=f(x-a)\in F[x]$ and we have $$ g(u+a) = f(u+a-a) = f(u) = 0. $$