# 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?

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 $$q(x)$$.