Nonlinear Equations in One Variable and Systems of Equations

$p(1) = 1 + a + b + 6 = 0 \Rightarrow a + b = -7$. Hmm, let's check $p(2) = 8 + 4a + 2b + 6 = 0 \Rightarrow 4a + 2b = -14 \Rightarrow 2a + b = -7$. Subtract: $a = 0$, $b = -7$. So $a + b = -7$. Wait, that doesn't match. Let me redo: $a = 0$, $b = -7$, so $a + b = -7$. Hmm, the answer should be different. Actually since the third root must be $r$ such that $1 \cdot 2 \cdot r = -6$, $r = -3$. Then $a = -(1+2+(-3)) = 0$, $b = (1)(2)+(1)(-3)+(2)(-3) = 2 - 3 - 6 = -7$. So $a + b = -7$. The correct answer should be $-7$, not $-6$. Let me state $-7$.