Vtyjkyuk

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?

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 $ninmathbbZ^+$.

So for $n=1:$

$f(x) = a_1x +a_0=0implies x=-fraca_0a_1$ 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 $forall ninmathbbZ^+$

By induction over $n$:

If $n=0$ then $f in F^times$ and $f(alpha)=fneq 0$ 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 $gin F[X]$ 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 $F$