<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] $$f(x)=\left(x^3+4 x^2+x-6\right) q(x)+a x^2+b x+c=\\=(x^2+2 x-3) (x+2)  q(x)+a x^2+b x+c\tag{1}$$
$$r(x)=a x^2+b x+c$$
We know that
$$\frac{f(x)}{x+2}$$
gives remainder $r=1$, which means that $f(-2)=1$
then we know that
$$\frac{f(x)}{x^2+2 x-3}$$
gives remainder $r=x+2$, which means that
$$f(x)=p(x)(x^2+2x-3)+x+2=p(x)(x-1)(x+3)+x+2$$
so we have $f(1)=3$ and $f(-3)=-1$
Now back to
$$f(x)=\left(x^3+4 x^2+x-6\right) q(x)+a x^2+b x+c=(x-1) (x+2) (x+3)q(x) +ax^2+bx+c$$
we plug $x=-2;\;x=1;\;x=-3$ and get
$$
\begin{cases}
4 a-2 b+c=1\\
a+b+c=3\\
9 a-3 b+c=-1\\
\end{cases}
$$
solution should be $$\left(a= -\frac{1}{3},b= \frac{1}{3},c= 3\right)$$
Remainder should be $$r(x)=-\frac{x^2}{3}+\frac{x}{3}+3$$
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