An algebra problem by A Former Brilliant Member

Given that $\begin{cases} r+s-a-b = 2 & & ...(1) \\ rs + a + b + 2 = 0 & \implies rs + a + b = -2 &...(2) \end{cases}$

Therefore,

$\begin{aligned} (1)+(2): \quad r+s+rs & = 0 \\ r(s+1) + s & = 0 \\ r(s+1) + s+1 & = 1 \\ (r+1)(s+1) & = \boxed{1} \end{aligned}$

(r+1)(s+1)=rs+r+s+1 r+s-a-b=2 which is the same as r+s-a-b-2=0 rs+a+b+2=0 we can say that rs+a+b+2+r+s-a-b-2=0 rs+r+s=0 sutitute 0 0+1=1