Question

Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x³ + ax² + bx – 12.

Answer 1

Let f(x) = x³ + ax² + bx – 12

x – 2 = 0

`\implies` x = 2 

x – 2 is a factor of f(x).

So, remainder = 0 

∴ (2)³ + a(2)² + b(2) – 12 = 0 

`\implies` 8 + 4a + 2b – 12 = 0 

`\implies` 4a + 2b – 4 = 0 

`\implies` 2a + b – 2 = 0  ...(1) 

x + 3 = 0

`\implies` x = –3     

x + 3 is a factor of f(x).

So, remainder = 0 

∴ (–3)³ + a(–3)² + b(–3) – 12 = 0 

`\implies` –27 + 9a – 3b – 12 = 0 

`\implies` 9a – 3b – 39 = 0 

`\implies` 3a – b – 13 = 0   ...(2)

Adding (1) and (2), we get, 

5a – 15 = 0 

`\implies` a = 3 

Putting the value of a in (1), we get,

 6 + b – 2 = 0 

`\implies` b = – 4