<s>[INST] $f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$) Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividing the polynomial with ($x + 1$)($x^2 + 1$)?
My attempt:
By Polynomial remainder theorem we know that 
$$f(x)=q_1(x)(x+1)+4$$
$$f(x)=q_2(x)(x^2+1)+(2x+3)$$
By putting $x=1$ we know that $f(-1)=4, f(i)=2i+3, f(-i)=-2i+3$
We want to find $r(x)$ such that: $$f(x)=(x+1)(x^2+1)q_3(x) + r(x) .$$
By applying the previous idea we know that $f(-1)=r(-1)=4$, but the same idea $r(i)=2i+3$ and $r(-i)=-2i+3$ but this is only three point and we need to determinant polynomial $r(x)$ of degree 3...
Please solve without modular arithmetic.  
 [/INST] Upon applying the law:  $\,ab\bmod ac = a(b\bmod c) = $ Mod Distributive Law we obtain
$(f\!-\!(2x\!+\!3))\bmod{(x^2\!+\!1)(x\!+\!1)}\,=\, (x^2\!+\!1){\Huge[}\dfrac{\overbrace{f\!-\!(2x\!+\!3)}^{\large 4\,-\,(2(\color{#c00}{\bf -1})+3)}}{\underbrace{x^2\!+\!1}_{\large (\color{#c00}{\bf -1})^2+1\ \ \ \ }}\underbrace{\bmod{x\!+\!1}}_{\Large x\ \equiv\ \color{#c00}{\bf {-}1}}{\Huge]}=\, (x^2\!+\!1)\left[\dfrac{3}{2}\right]$
So $\, f\equiv  2x\!+\!3 + \dfrac{3}2(x^2\!+\!1).\ $  Note that we did not need to solve any equations - just evaluate.
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