Question

Let R₁ and R₂ be the remainders when the polynomials x³ + 2x² – 5ax – 7 and x² + ax² – 12x + 6 are divided by (x + 1) and (x – 2) respectively. If 2R₁ + R₂ = 6, the value of a is :

(a) –2 

(b) 1 

(c) –1 

(d) 2

Answer 1

(d) 2

Let f(x) = x³ + 2x² – 5ax – 7

∴ R₁ = f(–1) = (–1)³ + 2. (–1)² – 5.a. (–1) –7 

= –1 + 2 + 5a –7 = 5a – 6 

g(x) = x³ + ax² – 12x + 6 

R₂ = g(2) = (2)³ + a.(2)² – 12.(2) + 6 

= 8 + 4a – 24 + 6 = 4a – 10 

Given, 2R₁ + R₂ = 6 ⇒ 2(5a – 6) + (4a – 10) = 6 

⇒ 10a – 12 + 4a – 10 = 6 

⇒ 14a – 22 = 6 ⇒ 14a = 28 ⇒ a = 2.