<s>[INST] Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x-3)(x+2)\\(x^2+2x-3)=(x-1)(x+3)\\
(x+2)\\
f(x)=(x^3+4x^2+x-6)g(x)+(Ax+B)\\
f(1)=A+B=x+2\\
f(-3)=-3A+B=x+2\\
f(-2)=-2A+B=1\\
\left\{\begin{matrix}
 A+B=x+2& x=-1\\ 
 -3A+B=x+2& B=1\\ 
 -2A+B=1& A=0
\end{matrix}\right.\\
r=(-1)(0)+1=1$$
Can the remainder really be $r=1$? My intuition says it's wrong.
EDIT:
$$f(x)=(x^3+4x^2+x-6)g(x)+(Ax^2+Bx+C)\\
f(1)=A+B+C=x+2\\
f(-3)=9A-3B+C=x+2\\
f(-2)=4A-2B+C=1\\
\left\{\begin{matrix}
 A+B+C=x+2& A=\frac{x+3}{6}\\ 
 9A-3B+C=x+2& B=\frac{3x+5}{6}\\ 
4A-2B+C=1 &C=\frac{4-2x}{6} 
\end{matrix}\right.\\
r=\frac{x+3}{6}x^2+ \frac{3x+5}{6}x+\frac{4-2x}{6}=\frac{x^3+6x^2+3x+9}{6}$$
 [/INST] Since:
$f(x) \equiv x+2 \ (mod \ x^2+2x-3)$
$f(x) \equiv 1 \ (mod \ x+2)$
$x^2+2x-3 \equiv -3 \ (mod \ x+2)$
We have that:
$f(x)=(-1/3x^2-2/3x+1)+(x+2)+k \ (x^3+4^2x+x-6)$
so the remainder is: $(-1/3x^2-2/3x+1)+(x+2)$
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