Answer
$b=2$
$c=-8$
They are unique
Work Step by Step
We have to find $b,c$ so that $p(x)=x^2+bx+c$ and $\lim\limits_{x \to 2}\left(\dfrac{p(x)}{x-2}\right)=6$.
In order for the limit to exist $p(x)$ should have a factor $(x-2)$. We must have:
$p(x)=(x-1)q(x)=(x-1)(x-a)$
$q(2)=6$
Determine $a$:
$2-a=6\Rightarrow a=-4$
The polynomial $p(x)$ is:
$p(x)=(x-2)(x+4)=x^2+4x-2x-8=x^2+2x-8$
Determine the constants $b,c$:
$b=2$
$c=-8$
Given the way we found them, the constants $b,c$ are unique.
