Question

If x³ + ax² + bx + 6 has x – 2 as a factor and leaves a remainder 3 when divided by x – 3, find the values of a and b.

Answer 1

Let f(x) = x³ + ax² + bx + 6

∴  x – 2 = 0 `\implies` x = 2

(2)³ + a(2)² + b(2) + 6 = 0

8 + 4a + 2b + 6 = 0

4a + 2b + 14 = 0

2(2a + b + 7) = 0   

2a + b + 7 = `0/2`

2a + b + 7 = 0

2a + b = –7     ...(i) 

∴ x – 3 = 0 `\implies` x = 3

(3)³ + a(3)² + b(3) + 6 = 3 

27 + 9a + 3b + 6 = 3

9a + 3b + 33 = 3

9a + 3b = 3 – 33

9a + 3b = –30

3(3a + b) = –30

3a + b = `(-30)/3`

3a + b = –10   ...(ii)

Subtracting (i) from (ii), we get,

2a + b = – 7
3a + b = – 10  
–    –        +     
– a = 3

∴ a = –3 

Substituting the value of a = –3 in (i), we get, 

2a + b = –7

2(–3) + b = –7

– 6 + b + 7 = 0

b = –7 + 6

∴ b = –1