Let $f(a, b, c)$ be a polynomial. What does $f(g(b, c), b, c)=0$ tell us about the factorization of $f(a, b, c)$? I was trying to factor $f(a, b,c)$ and by trial and error I discovered $f(-b-c, b, c) = 0$. I set $a= -b-c$ from which I got $a+b+c=0$, and I got that $(a+b+c)$ is a factor of $f(a, b, c)$.
Is it true in general that if $f(g(b, c), b, c)=0$, then $a - g(b, c)$ is a factor of $f$? 

My failed attempt:
I tried to prove this myself but I quickly saw that I need very clear definitions, which I don't posses at the moment. For example, I tried a proof by contradiction. I assumed that $f(g(b, c), b, c)=0$, and that  $f(a, b, c) = h_1(a, b, c) \cdot h_2 (a, b, c)$, and finally that neither of $h_1, h_2$ is $a-g(b, c)$. But of course, this wouldn't work. $h_1$ and $h_2$ can simply be expressions "containing" the factor $a - g(b, c)$. So I had to come up with a good definition for "unfactorability". But this was difficult too, because it's not clear to me wheather unfactorable$\implies$ can never be $0$ for any values. I saw that this question could lead me down a deep rabbit hole, so I decided to just post it here. 
 A: The trick is to write $f$ as a polynomial in $a$ first, with coefficients being polynomials in $b,c$. So you get something that looks like:
$$f(a,b,c)=(b^2+c)a^4+(b+c^5)a^2+(bc+b^2c^2)a+(b^4+c^4)$$
Then you "divide" by $a-g(b,c)$ in the usual way you'd divide a polynomial. So you are essentially treating polynomials with coefficients in $b,c$ as "numbers." (There is a notion for this in abstract algebra - we are looking at polynomial rings.)
After doing the polynomial long division, you get a remainder that is just a polynomial in only $b,c$, call it $r(b,c)$.
So you get $$f(a,b,c)=(a-g(b,c))q(a,b,c) + r(b,c)$$
for some polynomials $q,r$. But then $$f(g(b,c),b,c)=(g(b,c)-g(b,c)q(g(a,b),b,c)+r(b,c)=r(b,c)$$
you get that $r(b,c)=0$, and hence $f(a,b,c)=(a-g(b,c))q(a,b,c)$, which is the result you want.
A: Yes, that is true.
Your observation is that $f(a,b,c)=0$ whenever $a-g(b,c)=0$.
This allows us to apply the Nullstellensatz on the (principal) ideal generated by $a-g(b,c)$ to conclude that some power of $f(a,b,c)$ is divisible by $a-g(b,c)$.
However, $\mathbb R[a,b,c]$ is a unique factorization domain, and since $a-g(b,c)$ is clearly irreducible (its only term containing $a$ is $a$ itself), this means that if $a-g(b,c)$ divides a power of $f$, then it must divide $f$ itself.
