Given that (x + 1) is a factor of f(x)= (2x³ + ax² + 2bx + 1), then f(-1) = 0

[if is a factor of f(x)= ax² + bx + c then f( - α) = 0]

-2 + a-2b + 1 = 0
a - 2b - 1 = 0 …(i)

Also, 2a - 3b = 4

3b = 2a - 4

Now, put the value of b in Eq. (i), we get

3a - 2(2a - 4) - 3 = 0
3a - 4a + 5 = 0

a = 5

Now, put the value of a in Eq. (i), we get

5 - 2b - 1 = 0
2b = 4
b = 2

Hence, the required values of a and b are 5 and 2, respectively.