Finding remainder when unknown $f(x)$ is divided to $g(x)$

Question

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?

Answer 1

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 $q(x)$.