Prove that every equation of degree $n$ in $x$ has $n$ roots and no more. Prove that every equation of degree $n$ in $x$ has $n$ roots and no more.
My Attempt:
Let us suppose $f(x)=0$ where
$$f(x)=a_0 x^n + a_1 x^{n-1}+.....+a_{n-1} x+a_n$$
According to the Fundamental Theorem of Algebra, the equation $f(x)=0$ has a root, real or complex. Let this root be $\alpha_1$; then by the Factor Theorem, $x-\alpha_1$ is a factor of $f(x)$.
So, how do I proceed from here?
 A: We have $f(\alpha_1)=0$. Then we can write $f(x) = (x-\alpha_1)g(x)\quad$ where $g(x)$ is a polynomial with degree $n-1$.
So, $g(x)$ must have a root $\alpha_2$. Then we can write $f(x) =(x-\alpha_1)(x-\alpha_2)h(x)$ where $h(x)$ is a polynomial with degree $n-2$ etc...
In general we have a proposition that a polynomial $f(x)$ with degree $n$ has $n$ roots for $n\in\mathbb{Z^+}$.
So for $n=1:$
$f(x) = a_1x +a_0=0\implies x=-\frac{a_0}{a_1}$ is the only root. So true for $n=1$.
Assume true for $n$. Now we must prove true $n+1:$
Let $q(x)$ have degree $n+1$. So by the Fundamental Theorem of Algebra, there exists a root of $q(x)$ which we will call $\alpha_1$. Then we can write $q(x) = (x-\alpha_1)f(x)$ where $f(x)$ is some polynomial with degree $n$. From our inductive assumption, we have that $f(x)$ has $n$ roots. Therefore $q(x)$ has $n+1$ roots as required. Hence our statement holds $\forall n\in\mathbb{Z^+}$
A: By induction over $n$:
If $n=0$ then $f \in F^\times$ and $f(\alpha)=f\neq 0$ for every $\alpha \in F$.
That is $f$ has no roots in $F$.
Suppose $n>0:$
Suppose the claim is true for every polynomial of degree $n-1$.
If $f$ has not roots in $F$, we're done.
Else, it has root $\alpha \in F$, so $f=(X-\alpha)g,\ $ for some $g\in F[X]$ of degree $n-1$.
By induction hypothesis, $g$ has at most $n-1$ different roots in $F$.
It is easy to see that the roots of $f$ are exactly the roots of $g$ and $\alpha$.
Hence, $f$ has at most $n$ different roots in $F$
