Question

Find the values of m and n so that x – 1 and x + 2 both are factors of x³ + (3m + 1)x² + nx – 18.

Answer 1

Let f(x) = x³ + (3m + 1)x² + nx – 18 

x – 1 = 0 `\implies` x = 1 

x – 1 is a factor of f(x).

So, remainder = 0 

∴ (1)³ + (3m + 1)(1)² + n(1) – 18 = 0 

`\implies` 1 + 3m + 1 + n – 18 = 0  

`\implies` 3m + n – 16 = 0  ...(1) 

x + 2 = 0 `\implies` x = –2 

x + 2 is a factor of f(x).

So, remainder = 0 

∴ (–2)³ + (3m + 1)(–2)² + n(–2) – 18 = 0

`\implies` –8 + 12m + 4 – 2n – 18 = 0

`\implies` 12m – 2n – 22 = 0 

`\implies` 6m – n – 11 = 0   ...(2) 

Adding (1) and (2), we get, 

9m – 27 = 0 

m = 3 

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

3(3) + n – 16 = 0 

9 + n – 16 = 0  

n = 7