Question

If ax³ + bx² + x – 6 has (x + 2) as a factor and leaves a remainder 4, when divided by (x – 2), the value of a and b respectively are :

(a) 1, –2 

(b) 2, 1 

(c) 0, 2 

(d) 1, –1

Answer 1

(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.