(c) 0, 2
Let f (x) = ax3 + bx2 + \(x\) – 6
(x + 2) is a factor of f(x) ⇒ f(–2) = 0
∴ f(–2) = –8a + 4b – 2 – 6 = 0
⇒ –8a + 4b – 8 = 0 ⇒ –2a + b = 2 ...(i)
(x – 2) leaves a remainder 4, when dividing f(x) ⇒ f(2) = 4
∴f(2) = 8a + 4b + 2 – 6 = 4 ⇒ 8a + 4b – 8 = 0
⇒ 2a + b = 2 ...(ii)
From (i) and (ii) b = 2, a = 0.