If (x + 1) is a factor of 2x3 + ax2 + 2bx+l,

Solution:

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

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

$\Rightarrow \quad 2(-1)^{3}+a(-1)^{2}+2 b(-1)+1=0$

$\Rightarrow \quad-2+a-2 b+1=0$

$\Rightarrow \quad a-2 b-1=0$ $\ldots($ i)

Also, $2 a-3 b=4$

$\Rightarrow$ $3 b=2 a-4$

$\Rightarrow$  $b=\left(\frac{2 a-4}{3}\right)$

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

$a-2\left(\frac{2 a-4}{3}\right)-1=0$

$\Rightarrow \quad 3 a-2(2 a-4)-3=0$

$\Rightarrow \quad 3 a-4 a+8-3=0$

$\Rightarrow \quad-a+5=0$

$\Rightarrow \quad a=5$

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

$5-2 b-1=0$

$\Rightarrow \quad 2 b=4$

$\Rightarrow \quad b=2$

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