Nonlinear Equations in One Variable and Systems of Equations

$p(1) = 1 + a + b + 2 = 0$, so $a + b = -3$. $p(-2) = -8 + 4a - 2b + 2 = 0$, so $4a - 2b = 6$, i.e., $2a - b = 3$. Add: $3a = 0 \Rightarrow a = 0$, $b = -3$. Wait, $a + b = -3$ with $a = 0$ gives $b = -3$. But choice says $-1$. Recompute: $2a - b = 3$ with $a = 0$ gives $b = -3$. Hmm. Adjusting: actually with $a=0$, the polynomial is $x^3 + bx + 2$. With $b = -3$: $p(1) = 1 - 3 + 2 = 0$ ✓; $p(-2) = -8 + 6 + 2 = 0$ ✓. So correct $b = -3$.