How to prove whether a polynomial function is even or odd We know that a function is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. With this reasoning is it possible to prove that a polynomial function such as  $f(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + ...+a_{2}x^2 + a_{0}$  is even or odd? 
What do you suggest? How do we get started? 

Two polynomials have the same value at every real number if and only if they are identical (exact same coefficients in each and every power of $x$); this follows because a polynomial of degree $n\gt 0$ can have at most $n$ roots. If $f(x)$ and $g(x)$ are the same at every value of $x$, then $f-g$ has infinitely many roots, and so must be the zero polynomial.
So write
$$f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x + a_0.$$
Then
$$\begin{align*}
f(-x) &= a_n(-x)^n + x_{n-1}(-x)^{n-1} + \cdots + a_1(-x) + a_0\\
&= a_n (-1)^nx^n + a_{n-1}(-1)^{n-1}x^{n-1} + \cdots + a_1(-1)x + a_0\\
&= \left((-1)^na_n\right)x^n + \left((-1)^{n-1}a_{n-1}\right)x^{n-1} + \cdots + \left(-a_1\right)x + a_0.
\end{align*}$$
Now, $f(x)$ is even if and only if $f(x) = f(-x)$. That means that we need
$$a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = (-1)^nx^n + (-1)^{n-1}a_{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$
That, in turn, means we need
$$a_n = (-1)^{n}a_n,\quad a_{n-1}=(-1)^{n-1}a_{n-1},\ldots,\quad a_1 = -a_1, \quad a_0=a_0.$$
If $k$ is even, then $(-1)^k = 1$, so we automatically get $a_k=(-1)^ka_k$; that just says we need $a_k=a_k$, which is always true. If $k$ is odd, then $(-1)^k = -1$, so we need $a_k=-a_k$. This can only happen if $a_k=0$.
So $f(x)$ is an even function exactly when all odd terms to have coefficient $0$. So it must be a polynomial in which the only powers of $x$ that "show up" are even powers of $x$ (including $x^0$ which gives the constant term).
For $f(x)$ to be an odd function, we need $-f(x)=f(-x)$. That means that we need:
$$-\Bigl(a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0\Bigr) = (-1)^nx^n + (-1)^{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$
That, in turn, means we need
$$-a_n = (-1)^{n}a_n,\quad -a_{n-1}=(-1)^{n-1}a_{n-1},\ldots\quad, -a_1 = -a_1, \quad -a_0=a_0.$$
This time, if $k$ is odd, then we are asking for $-a_k = -a_k$ to be true, which it always is; and if $k$ is even we are asking for $-a_k = a_k$, which is true if and only if $a_k=0$. 
So $f(x)$ is an odd function exactly when all the even coefficients are zero; that is, the only powers that "show up" are odd powers of $x$.
