What is the value of b+c if x^2+bx+c>0 only when x\in (-\infty, -2)\cup(3,\infty)?
+2666 \( x\in (-\infty, -2)\cup(3,\infty)\)
So we have \((-2)^2 -2b + c = 0\), meaning \(-2b + c = -4\)
We also have \((3)^2 + 3b + c = 0\), meaning \(3b + c = -9\)
Solving the system gives us \(b = -1\) and \(c = -6\).
Thus, \(b + c = -1 - 6 = \color{brown}\boxed{-7}\)
