<s>[INST] What is the remainder of polynomial $f(x) g(x)$ divided by $x^{2} + x - 6$? $F(x),g(x)$ be polynomials. $f(x)$ divided by $x^{2}-4$ has remainder $ax+a$. $g(x)$ divided by $x^{2}-9$ has remainder $ax+a-5$. It is known that the remainder 
$$ Remainder \: f(x) \: divided \: by \: x+2 = Remainder \: g(x) \: divided \: by \: x-3  $$
and $f(-3)=g(2)=-2$.
What is the remainder of $f(x)g(x)$ divided by $x^{2}+x-6$?

Attempt:
By the first known information we get $f(2) = 3a, f(-2) = -a,  g(3) = 4a-5, g(-3) = -2a-5$
Then using the 2nd known information we get
$$ -a = 4a - 5 \implies a = 1 $$
So we have $f(x)$ is of the form
$$ f(x) = (x^{2} - 4) F(x) + (x+1) $$
where $F(x)$ is a polynomial. Also
$$ g(x) = (x^{2}-9)G(x) + (x-4) $$
Now we have to find the form of 
$$ f(x) g(x) = (x+3)(x-2) H(x)  + R(x) $$
with $f(2)g(2) = -6 =R(2)$, and $ f(-3) g(-3) = 14 = R(-3)$
 [/INST] Since you're dividing by a quadratic polynomial, the remainder has to be a linear polynomial, because its degree must be strictly less than that of the divisor. So you can write your remainder as $R(x)=px+q$. Knowing that $R(2)=-6$ and $R(-3)=14$, you can set up and solve a system of two equations with two unknowns $p$ and $q$.
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