Question 5: Given $A=\left\{x \mid x^{3}+3 x^{2}+2 x>0\right\}, B=\left\{x \mid x^{2}+a x+b \leq 0\right\}$, and $A \cap B=$ $\{x \mid 0<x<2\}$, then $a+b=$ $\qquad$ _.
Question 5, Answer: Since $x^{3}+3 x^{2}+2 x=x(x+1)(x+2)$, we have $A=(-2,-1) \cup(0,+\infty)$. Let the two roots of the equation $x^{2}+a x+b=0$ be $x_{1} \leq x_{2}$, then $B=\left[x_{1}, x_{2}\right]$. Since $A \cap B=\{x \mid 0-2\}$, it follows that $x_{2}=2, x_{1}=-1$, hence $a=-\left(x_{1}+x_{2}\right)=-1, b=x_{1} x_{2}=-2$, thus $a+b=-3$.