<s>[INST] Finding remainder when unknown $f(x)$ is divided to $g(x)$ When $f(x)$ is divided by $x - 2$ and $x + 3$, the remainders are 5 and -1, respectively. Find the remainder when $f(x)$ is divided by $x^2 + x - 6$
My method:
Since $x - 2$ and $x + 3$ are linears, dividing by quadratic will leave linear remainders. 
Using the remainder theorem:
$$ f(x) = g(x) q(x) + r(x) $$
Where $g(x)$ is a divisor, $q(x)$ is a remainder, and $r(x)$ is a remainder
I let $ax + b$ here be the remainder.
So:
\begin{align}
f(x) &= g(x) (x - 2)(x + 3) + ax + b\\
f(2) &= g(2) (0)(5) + 2a+ b\\
f(-3) &= g(-3) (-5)(0) - 3a + b\\
\\
&5 = 2a + b\\
&-1 = -3a + b\\
\\
&...\\
\\
&a = \frac{6}{5}, b = \frac{13}{5}
\\
\end{align}
Then the remainder is $ax + b = \frac{6}{5}x + \frac{13}{5} $
Is it possible to find the $f(x)$ out of remainders?
 [/INST] No, there are infinite $f(x)$ which satisfy the given conditions:
$$f(x)=q(x)(x^2 + x - 6)+\frac{6}{5}x + \frac{13}{5}$$
with any polynomial $q(x)$.
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