We have,

p (x) = ax³ + bx² + x – 6

It is given in the question that, (x + 1) is a factor of p (x)

x + 2 = 0

x = - 2

∴ f (-2) = 0

a (-2)³ + b (-2)² + (-2) – 6 = 0

- 8a + 4b – 2 – 6 = 0

- 8a + 4b – 8 = 0

- 4 (2a – b + 2) = 0

2a – b + 2 = 0 (i)

Also, it is given in the question that when p (x) is divided by (x – 2) then it leaves a remainder 4

∴ p (2) = 4

a (2)³ + b (2)² + (2) – 6 = 4

8a + 4b + 2 – 6 = 4

8a + 4b – 4 - 4= 0

4 (2a + b – 2) = 0

2a + b – 2 = 0 (ii)

Now, adding (i) and (ii) we get:

2a – b + 2 + 2a + b – 2 = 0

4a = 0

a = 0

Putting the value of a in (ii), we get:

2 (0) + b – 2 = 0

b – 2 = 0

b = 2

Hence, it is proved that the value of a is 0 and that of b is 2