Choose any $a \in K$ and consider the evaluation homomorphism $\operatorname{ev}_a: F[x] \rightarrow F[a]$. Now apply the isomorphism theorems. Certainly, $\operatorname{Im}(\operatorname{ev}_a) = F[a]$.

Now, notice that $\ker\operatorname(ev_a)$ cannot be trivial. Otherwise, we'd have $F[x] \cong F[a]$, which is absurd since the latter is a field and the former is not. Therefore, the homomorphism has nontrivial kernel, which is precisely those polynomials with $a$ as a root.

Example: I can prove that $\sqrt[n]{2}$ is not rational for all $n > 2$.

Suppose $\sqrt[n]{2} = \frac{p}{q}$. Then we'd have $2 = \frac{p^n}{q^n}$. Rewriting this, we have $2q^n = p^n$, that is, we have $q^n + q^n = p^n$. This is false by Fermat's last theorem.

harmonics..wave equations I am doing this problem in my notebook right now
http://math.stackexchange.com/questions/1013087/solving-an-inhomogeneous-wave-equation-with-free-end-boundary-conditions

I keep getting $$ \arctan \frac{1}{a} - \int_0^{\pi/2} \frac{a \sin x - 1}{a^2 - 2 a \sin x + 1}\log \cos x\,\mathrm{d}x $$
But by numerical calculations this seem wrong. Where did I make a mistake?