Question

The expression 2x³ + ax² + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.

Answer 1

Let f(x) = 2x³ + ax² + bx – 2

2x – 3 = 0 `\implies x = 3/2` 

On dividing f(x) by 2x – 3, it leaves a remainder 7.

∴ `2(3/2)^3 + a(3/2)^2 + b(3/2) - 2 = 7`  

`27/4 + (9a)/4 + (3b)/2 = 9`  

`(27 + 9a+ 6b)/4 = 9` 

27 + 9a + 6b = 36  

9a + 6b – 9 = 0 

3a + 2b – 3 = 0   ...(1)

x + 2 = 0 `\implies` x = –2

On dividing f(x) by x + 2, it leaves a remainder 0. 

∴ 2(–2)³ + a(–2)² + b(–2) – 2 = 0 

–16 + 4a – 2b – 2 = 0 

4a – 2b – 18 = 0  ...(2)

Adding (1) and (2), we get,   

7a – 21 = 0 

a = 3 

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

3(3) + 2b – 3 = 0 

9 + 2b – 3 = 0 

2b = –6 

b = –3