## Calculus: Early Transcendentals (2nd Edition)

$b=2$ $c=-8$ They are unique
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.