### Instruction: What is the value of $b+c$ if $x^2+bx+c>0$ only when $x\in (-\infty, -2)\cup(3,\infty)$?### Response: I notice that the inequality is a quadratic expression, and I know that quadratics change sign at their roots.
So, if $x^2+bx+c>0$ only when $x\in (-\infty, -2)\cup(3,\infty)$, that means the roots of the quadratic are $x=-2$ and $x=3$.
I can use this information to find $b$ and $c$, since I know that the sum of the roots is equal to $-b/a$ and the product of the roots is equal to $c/a$, where $a$ is the leading coefficient of the quadratic.
In this case, $a=1$, so I can simplify these formulas to $-2+3=-b$ and $-2\cdot 3=c$.
Solving for $b$ and $c$, I get $b=-1$ and $c=-6$.
Therefore, the value of $b+c$ is $-1+(-6)=-7$.