Solve it!

$2x^4-9x^3+14x^2-9x+2=(x-1)^2(2x-1)(x-2)$ . Therefore, the distinct integer roots are $1$ and $2$ , so the answer is $\boxed{3}$ .

The sum of coefficients is $0$ . Hence it can be inferred that $x = 1$ is one of the roots.

By using Synthetic Division, we get

$2x^4 - 9x^3 + 14x^2 - 9x + 2 = (x-1)(2x^3 - 7x^2 + 7x - 2) = (x-1)(p(x))$

The sum of coefficients of $p(x)$ is also $0$ . Hence, $x =1$ is a root of $p(x)$ as well as of the original given polynomial.

Again by Synthetic Division,

$2x^4 - 9x^3 + 14x^2 - 9x + 2 = (x-1)(x-1)(2x^2 - 5x + 2) = (x-1)^2 (2x^2 - 5x + 2)$

$\Rightarrow 2x^2 - 5x + 2 = 2x^2 - 4x - x + 2 = 2x(x-2) -1(x-2) = (2x-1)(x-2)$

$\Rightarrow 2x^4 - 9x^3 + 14x^2 - 9x + 2 = (x-1)^2(2x-1)(x-2)$

Sum of the distinct roots is $\color{#3D99F6}{\boxed{3}}$ .