Divisible or not?

First, I should state the Remainder Theorem. According to it if $(x-b)$ divides a polynomial P(x), then P(b)=Remainder after division and if $(x-b)$ is a factor of a polynomial P(x), then P(b)=0 since then there is no remainder.

So, the polynomial P(x) here is divided by $(x-2)$ , so P(2)=0

$P(2)=0$

$\implies 2a\times 2^{3} - 4\times 2^{2} + 2a - 2a=0$

$\implies 2a\times 8 - 4\times 4=0$

$\implies 16a-16=0 \implies 16a=16 \implies a=\boxed{1}$

P(x) is the remainder of the polynomial. P(x) = 0, since (x - 2) is a factor of the polynomial 2ax³ - 4x² + ax - 2a.
To find the value of x, we have to find the root of the polynomial factor -- (x - 2).

Assume that x - 2 = 0.
Therefore, x = 2.
Now substitute the value of x in the equation to find 'a': P(x) = 2ax³ - 4x² + ax - 2a
P(2) = 2a(2)³ - 4(2)² + a(2) - 2a
0 = 2a(8) - 4(4) + 2a - 2a
0 = 16a - 16
16 = 16a
16a = 16
a = 16/16
Therefore, a = 1.