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.
 A: Hint $ $ We can read off a CRT solution from the Bezout equation for the gcd of the moduli, viz. $$\bbox[5px,border:1px solid #c00]{\text{$\color{#90f}{\text{scale}}$ the Bezout equation by the residue difference - then ${\rm \color{#c00}{re}\color{#0a0}{arrange}}$}}$$
$$\begin{align}
{\rm if}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\begin{array}{rr} &f\equiv\,  f_g\pmod{\!g}\\ &f\equiv\, f_h\pmod{\! h} \end{array}\ \ {\rm and}\ \ \gcd(g,h) = 1\\[.4em]
{\rm then}\ \ \  f_g - f_h\, &=:\ \delta\qquad\qquad\ \ \rm residue\ difference \\[.2em] 
\times\qquad\quad\ \ \ 1\ &=\ \ a g\, +\, b h\quad\ \rm Bezout\ equation\ for \ \gcd(g,h) \\[.5em]\hline
\Longrightarrow\ \,f_g\, \color{#c00}{-\, f_h}\, &= \color{#0a0}{\delta ag} + \delta bh\quad\ \rm product\ of \ above\ (= {\color{#90f}{scaled}}\ Bezout)\\[.2em]
\Longrightarrow \underbrace{f_g \color{#0a0}{- \delta ag}}_{\!\!\!\large \equiv\ f_{\large g}\! \pmod{\!g}}\! &= \underbrace{\color{#c00}{f_h} + \delta bh}_{\large\!\! \equiv\ f_{\large h}\! \pmod{\!h}}\ \ \ \underset{\large {\rm has\ sought\ residues}\phantom{1^{1^{1^{1^1}}}}\!\!\!}{\rm \color{#c00}{re}\color{#0a0}{arranged}\ product}\rm\! = {\small CRT}\ solution\end{align}  $$
More generally: $ $  if the gcd $\,d\neq 1\,$ then it is solvable $\iff d\mid f_g-f_h\,$ and we can use the same method we used below for $\,d=\color{#c00}2\!:\,$ scale the Bezout equation by $\,(f_g-f_h)/d = \delta/d.\,$ Since  $\,\color{#c00}2\,$ is invertible in the OP, we could have scaled the Bezout equation by $\,1/2\,$ to change $\,\color{#c00}2\,$ to $\,1,\,$ but not doing so avoids (unneeded) fractions so  simplifies the arithmetic.
In our specific problem we have the major simplification that  the Bezout equation is obvious being simply the moduli difference $ =\color{#c00}2$
hence $\ \ \smash[t]{\overbrace{\color{0a0}{6x\!-\!1}-\color{#90f}{(2x\!+\!1)}}^{\rm residue\ difference}} = \overbrace{(2x\!-\!1)}^{\!\text{scale LHS}}\,\overbrace{\color{#c00}2 = (\color{0a0}{x^2\!+\!6}-\color{#0a0}{(x^2\!+\!4)}}^{{\overbrace{\textstyle\color{#c00}2\, =\, x^2\!+\!6-(x^2\!+\!4)_{\phantom{|_{|_i}}}\!\!\!\!}^{\Large \text{Bezout equation}}}})\overbrace{(\color{#0a0}{2x\!-\!1})}^{\text{scale RHS}},\ $ which rearranged
yields $\ \ \underbrace{\color{}{6x\!-\!1 - (\color{#0a0}{2x\!-\!1})(x^2\!+\!6)}}_{\large 
 \equiv\ \ 6x\ -\ 1\ \pmod{x^2\ +\ 6}\!\!\!}\, =\,  \underbrace{\color{#90f}{2x\!+\!1} -\color{#0a0}{(2x\!-\!1)(x^2\!+\!4)}}_{\large \equiv\ \ 2x\ +\ 1\ \pmod{x^2\ +\ 4}\!\!\!} =\,r(x) =\, $  CRT solution.

Remark $ $ If ideals and cosets are familiar then the above can be expressed more succinctly as
$$  \bbox[12px,border:2px solid #c00]{f_g\! +\! (g)\,\cap\, f_h\! +\! (h) \neq \phi \iff f_g-f_h \in (g)+(h)}\qquad$$
A: 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:
$$f(x) \equiv \underbrace{(2x+1)(\ldots) + (6x-1)(\ldots)}_{r(x)} \pmod{(x^2+4)(x^2+6)}.$$
