APPROACH #1: Algebra
Given: \(\frac{2z^2(z-1)+z-z^2}{z(z-1)}\)

Rearrange the last two terms in the numerator: \(\frac{2z^2(z-1)-z^2+z}{z(z-1)}\)

Factor out \(-z\) from the last two terms of numerator: \(\frac{2z^2(z-1)-z(z-1)}{z(z-1)}\)

Rewrite the numerator as follows: \(\frac{(2z^2-z)(z-1)}{z(z-1)}\)

Simplify: \(\frac{2z^2-z}{z}\)

Finally, we can divide numerator and denominator by \(z\) to get: \(2z-1\)

Answer: D


APPROACH #2: Test a value of z
Since we are looking for an expression that's equivalent to \(\frac{2z^2(z-1)+z-z^2}{z(z-1)}\), let's first evaluate this expression for a certain value of \(z\), and then look for an answer choice that has the same value for that value of \(z\)

Let's see what happens when we plug \(z = 2\) into the given expression:
\(\frac{2z^2(z-1)+z-z^2}{z(z-1)} = \frac{2(2^2)(2-1)+2-2^2}{2(2-1)}\) \(= \frac{8+2-4}{2}\) \(= \frac{6}{2} = 3\)

So, the given expression evaluates to equal \(3\) when \(z = 2\).

We can now plug \(z = 2\) into each answer choice to see which one evaluates to \(3\)

(A) \(2(2^2) − 2 =6\). ELIMINATE
(B) \(2(2) + 1=5\). ELIMINATE
(C) \(2(2)=4\). ELIMINATE
(D) \(2(2) − 1=3\). KEEP
(E) \(2 − 2=0\). ELIMINATE

Answer: D