Ths isn't the proper Fundamental Theorem of Arithmetic (which includes complex roots), but just the maximum number of real roots for polynomial $f(x)$ with real coefficients... which is $deg(f)$.
I believe there isn't a High School-level proof of the full theorem, but is there one of this simpler theorem?
I think there should be a very simple proof using the Factor Theorem:
$$f(a)=0 \iff (x-a) \mid f(x)$$
The only way to have an extra zero is to have an extra factor of $(x-a)$, which increases the degree by one.
Because an extra factor might be the same as one already present, forming a multiple root or zero, the number of distinct roots may be fewer than this. So it is only a maximum.∎
However, I'm not sure how to make this rigorous, or how to show it covers all possible polynomials (i.e. all polynomials can be built up from $(x-a)$ factors - maybe it follows from the Factor Theorem?).
Or maybe there's a completely different, simpler approach?
BTW: I am interested in it as a way to understand the Schwartz-Zippel lemma for Polynomial Identity Testing (a probabilistic algorithm: subtracting the polynomials to get zero if identical, then use the maximum zeros/roots to calculate probability of guessing roots - false positives). The relevant part to this question is:
A polynomial of one variable of degree {d} can vanish at only {d} points without being identically zero The Curious History of the Schwartz-Zippel Lemma
Also at the wikipedia article
a polynomial of degree d can have no more than d roots.
Because the guesses can be made using only integers (I think modulo, making them fields), it seems to me that the simpler theorem for real polynomials should be enough...
If $P(x)$ is a polynomial such that $P(a)=0$ then it is only possible when it can be expressed as $(x-a)F(x)$ because when we subsitute $x=a$ we get $0$
The second statement is also obvious,
Let $P(x)$ have $p$ roots then it may be written as $(x-p_1)(x-p_2)...(x-p_p)$ which is a polynomial of degree p.