If two factors of $2x^3-hx+k$ are $x+2$ and $x-1$, the value of $|2h-3k|$ is
$\text{(A) } 4\quad \text{(B) } 3\quad \text{(C) } 2\quad \text{(D) } 1\quad \text{(E) } 0$
From the Remainder Theorem, we have $2(-2)^3 - h(-2) + k = 0$ and $2(1)^3 - h(1) + k = 0$.  Simplifying both of those equations gives $-16 + 2h + k = 0$ and $2 - h + k = 0$.  Since $k = 16 - 2h$ and $k = h - 2$, we set those equal to get:
$16 - 2h = h - 2$
$3h = 18$
$h = 6$
This gives $k = 4$ when substituting back into either of the two equations, and $|2h - 3k| = 0$, which is answer $\fbox{E}$.