Remainder of polynomial product, CRT solution via Bezout Given:
$$f(x) \pmod{x^2 + 4} = 2x + 1$$
$$f(x) \pmod{x^2 + 6} = 6x - 1$$
Define r(x) as:
$$f(x) \pmod{(x^2 + 4)(x^2+6)} = r(x)$$
What is $r(4)$?

The 3 equations can be restated as quotient · divisor + remainder:
$$f(x) = a(x)(x^2 + 4) + 2x + 1 $$
$$f(x) = b(x)(x^2 + 6) + 6x - 1 $$
$$f(x) = c(x)(x^2 + 4)(x^2 + 6) + r(x) = c(x)(x^4 + 10x^2 + 24) + r(x) $$

Note this isn't homework, and there are several different methods that can be used to solve this, one of which produces an f(x) based on the 2 given remainders, two of which produce r(x) without having to determine f(x), and a slight variation that produces r(4). I've looked at other polynomial remainder questions here at SE, but those did not involve all of the methods that I'm aware of that can be used to solve this particular problem, so I thought it might be interesting for others here at SE. Some, but not all of the methods are related to Chinese remainder theorem, so I wasn't sure if should also tag this question with Chinese remainder theorem. I found this problem at another forum site, so I'm not sure of the origins of this particular problem.

Hint
Observe that $\gcd(x^2+4, x^2+6)=1$ and 
$$\frac{1}{2}(x^2+6)-\frac{1}{2}(x^2+4)=1.$$
Now apply Chinese remainder theorem to the system
\begin{align*}
f(x) & \equiv 2x+1 \pmod{x^2+4}\\ 
f(x) & \equiv 6x-1 \pmod{x^2+6} 
\end{align*}
To get something like:
$$f(x) \equiv \underbrace{(2x+1)(\ldots) + (6x-1)(\ldots)}_{r(x)} \pmod{(x^2+4)(x^2+6)}.$$
