<s>[INST] Given 2 roots and a remainder of a cubic polynomial, find the remainder of $f(3)$ 
$f(x)$ is a cubic polynomial and the coefficient of $x^3$ is $k$. If the polynomial has exactly $2$ roots and these roots are $2$ and $-5$ and $f(1) = -48$, find the possible values of $f(3)$.

I've gotten part of the cubic equation to be $(kx+a)(x^2+3x-10)$, then I let $f(x) = (kx+a)(x^2+3x-10)$, substitute in $f(1)$ to end up with $-6k-a=-48$, and I am stuck from here onward. Any hint will be appreciated.
Thanks!
 [/INST] We have also $2k+a=0$ or $-5k+a=0$.
In the first case $f(x)=k(x-2)^2(x+5)$, which gives $-48=6k$ or $k=-8$ and
$f(3)=-8\cdot1^2\cdot8=-64$.
In the second case $f(x)=k(x+5)^2(x-2)$, which gives $-48=-36k$ or $k=\frac{4}{3}$ and
$f(3)=\frac{4}{3}\cdot8^2\cdot1=\frac{256}{3}$.
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