Remainder of polynomial product, CRT solution via Bezout

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{|_|}}\!\!\!}^{\Large \text{Bezout equation}}}})\overbrace{(\color{#0a0}{2x\!-\!1})}^{\text{scale RHS}},\ $ which rearranged

yields $\ \ \underbrace{\color{}{6x\!-\!1 - (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.

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Remark $ $ If ideals and cosets are familiar then the above can be expressed more succinctly as

$$ \bbox[12px,border:1px solid #c00]{f_g\! +\! (g)\,\cap\, f_h\! +\! (h) \neq \phi \iff f_g-f_h \in (g)+(h)}\qquad$$

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: $$f(x) \equiv \underbrace{(2x+1)(\ldots) + (6x-1)(\ldots)}_{r(x)} \pmod{(x^2+4)(x^2+6)}.$$