Why does splitting a polynomial into even and odd degree terms results in polynomials taking as input $x^2$? The question came up under comments to this YT presentation on FFT as a point taken as self evident, and Math SE seems like a natural place to answer it.
The idea is to evaluate a polynomial of degree $n$ at $\pm x_1, \pm x_2, \dots, \pm x_{n/2}$ points.
The example given is $P(x)= 3x^5 + 2x^4 + x^3 + 7 x^2 + 5x +1$ split as $P(x) = (2x^4+7x^2+ 1 x^0) + x (3x^4 + x^2 + 5).$
The next relabeling is confusing:
$$P(x) = (2x^4+7x^2+ 1 x^0) + x (3x^4 + x^2 + 5)=P_{\text{even}}(x^2) + xP_{\text{odd}}(x^2).$$
These even and odd degree polynomials are functions of $x^2$ and hence polynomials of degree $2.$
Here is the transcription of the YT presentation at this time:

... a bonus fact is that since these even and odd polynomials are functions of $x^2$ they are polynomials of degree $2,$ which is a much simpler problem than our original degree $5$ polynomial.

This is totally in line with the points made in comments about the author making reference to the specific example polynomial above. However, it does go on:

Generalizing these observations if we have an $n-1$ degree polynomial that we want to evaluate at $n$ points, we can split the polynomial into even and odd terms with these two smaller polynomials now having degree $n/2 - 1.$

This is also pointed out in the comments, and makes complete sense.
Evidently, the claim is not that $x^4$ is of degree $2,$ or anything outrageous like that. The observation is that this decomposition into even and odd polynomials can be regarded as functions of $x^2$ to make sure it matches the appropriate coefficients, as explained also in this MIT lecture at this juncture:
If the generic polynomial in which the idea is to evaluate sample points is
$$P(x) = a_nx^n+ a_{n-1}x^{n-1}+\cdots+ a_1 x^1+a_0 x^0$$
Then this can be split into
$$P_{\text{even}}(x^2)=\sum_{k=0}^{n/2} a_{2k}(x^2)^k$$
and
$$P_{\text{odd}}(x^2)=\sum_{k=0}^{n/2-1} a_{2k+1}(x^2)^k$$
but to match the coefficients with the powers of $x,$ these even and odd polynomials (of half the degree of the original polynomial) have to be evaluated at $x^2$ when combined as
$$P(x) =P_{\text{even}}(x^2) + xP_{\text{odd}}(x^2).$$
For instance, considering the even polynomial, at $k=n/2,$ the input $x$ would be raised to $(x^2)^{n/2}=x^n$ and this would be matched with the coefficient $a_{n}.$ On the odd side, at $k=n/2-1,$ the coefficient to be matched with $x^{2(n/2-1)}=x^{n-2}$ (of course multiplied by the $x$ in front, yielding $x^{n-1})$ is $a_{2(n/2-1)+1}=a_{n-1}.$
 A: 
These even and odd degree polynomials are functions of $x^2$ and hence
polynomials of degree $2.$

NO.
If $P(x)$ has degree $d$, then $P(x^2)$ has degree $2d$.

In particular, if $$P(x)=a_nx^n + a_{n-1}x^{n-1}+\cdots + a_0,$$
then
$$P(x^2)=a_nx^{2n} + a_{n-1}x^{2n-2}+\cdots + a_1x^2 + a_0$$

What the solution you cite is doing is using the following fact:

If $P$ is a polynomial such that all coefficients of odd powers of $x$ are zero, then there exists some polynomial $Q$ such that, for all $x$, $P(x)=Q(x^2)$ (in other words, $P=Q\circ S$ where $S$ is the polynomial defined as $S(x)=x^2$.

This fact can very easily be shown to be correct by just writing the polynomial $P$ out.

Example:
What the step does is it decomposes the polynomial. To make it more clear, look at a simple example of a fourth degree polynomial. Let's say
$$P(x) = x^4+4x^3 + 9x^2-3x-19\tag{1}$$
Then, the presentation you link simply rewrites the polynomial like so:
$$P(x)=(x^4+9x^2-19) + x(4x^2-3)$$
Now, they define two polynomials, $P_e$ and $P_o$, like so:
$$P_e(x)=x^2+9x-19\\
P_o(x) = 4x-3\tag{2}$$
These definitions are such that the following equality holds:
$$P(x) = P_e(x^2) + x\cdot P_o(x^2).\tag{3}$$
You can easily check that this equality holds by just substituting the values and writing everything out.
