Polynomial division. If an f(x) is divided by x²+4, has 2x+3 of remainder.
If an f(x) is divided by x²+6, has 6x-1 of remainder.
If an f(x) is divided by (x²+4)(x²+6), has S(x) of remainder, then find S(4)!
I have this.
I wrote it in polynomial formula.
$f(x)=(x²+4).R(x)+(2x+3)$
And yes, i have x²=-4. 
But, i had to express the 2x+3 function into the (x²) one. So, basically if we have p(x)=2x+3, then p(x²)=2x²+3. and i can put -4 to the x, and i got : 
f(-4)=2(-4)+3 = -5 ( and the same thing happened with x²+6 ). Is that true so far?
Sorry for bad english.
 A: Use the Chinese remainder theorem for polynomials:
A Bézout's relation between the coprime polynomials $x^2+4$ and $x^2+6$ is
$$\tfrac12(x^2+6)-\tfrac12(x^2+4)=1,$$
hence the solution of the simultaneous congruences
$\quad\begin{cases}f(x)\equiv 2x+3\mod x^2+4\\f(x)\equiv 6x-1\mod x^2+6\end{cases}\quad$
is $$S(x)\equiv\tfrac12\bigl[(2x+3)(x^2+6)-(6x-1)(x^2+4)\bigr]\mod(x^2+4)(x^2+6),$$so that $\;S(4)=-109$.
A: Approach using complex numbers.
Notice that $x^2+4=0$ when $x=\pm 2i$ and $x^2+6=0$ when $x=\pm i\sqrt{6}$ since we have that $$f(x)=(x^2+4)R(x)+2x+3$$
From this we have that $$f(2i)=4i+3\\f(-2i)=-4i+3$$
Also we have that
$$f(x)=(x^2+4)Q(x)+6x-1\\f(i\sqrt{6})=6i\sqrt{6}-1\\f(-i\sqrt{6})=-6i\sqrt{6}-1$$
From this we have that
$$f(x)=(x^2+4)(x^2+6)T(x)+ax^3+bx^2+cx+d\\f(2i)=-8ai-4b+2ci+d\\f(-2i)=8ai-4b-2ci+d\\f(i\sqrt{6})=-6a\sqrt{6}i-6b+ci\sqrt{6}+d\\f(-i\sqrt{6})=6a\sqrt{6}i-6b-ci\sqrt{6}+d$$
Now $$f(2i)+f(-2i)=2d-8b=6\\ f(i\sqrt{6})+f(-i\sqrt{6})=2d-12b=-2$$ from this we have that $b=2$ and $d=11$. We also have that
$$f(2i)-f(-2i)=4ci-16ai=8i\\c-4a=2\\f(i\sqrt{6})-f(-i\sqrt{6})=2ci\sqrt{6}-12ai\sqrt{6}=12\sqrt{6}i\\c-6a=6$$
Solving this we get $a=-2$ and $c=-6$
This implies that $S(x)=-2x^3+2x^2-6x+11$ and $S(4)=-128+32-24+11=-109$
