Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).
First, carry out the division:
f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)
Observe that
2z ² - 3z + 1 = (2z - 1) (z - 1)
so you can separate the rational part of f(z) into partial fractions. We have
(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)
8z - 2 = a (z - 1) + b (2z - 1)
8z - 2 = (a + 2b) z - (a + b)
so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.
So we have
f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)
or
f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))
Recall that for |z| < 1, we have

Replace z with 1/z to get

so that by substitution, we can write

Now condense f(z) into one series:




So, the inverse Z transform of f(z) is
.