Recall the pattern $a^2 - b^2 = (a - b)(a + b)$. Apply this to $4z^2 - 25$ and see what product of two binomials you get.

Once you factor the numerator, rewrite the whole fraction and look for a factor that appears in both the numerator and the denominator. What is left after you cancel that common factor (keeping in mind $z \neq -\tfrac{5}{2}$)?

In Desmos, type y = (4x^2 - 25) / (2x + 5) to represent the original expression (Desmos uses x instead of z, which is fine for checking equivalence).

For each answer choice, enter a new line in Desmos, such as y = 2x + 5, y = (2x - 5)(2x + 5), and y = 4x^2 - 25, and also y = 2x - 5. Visually compare each graph to the graph of the original expression.

The correct choice will have a graph that overlaps the graph of the original expression at all x-values where the original expression is defined (there will be a missing point at $x = -2.5$ for the original). Pick the option whose graph matches this behavior.

The numerator is $4z^2 - 25$. Notice that $4z^2$ is $(2z)^2$ and $25$ is $5^2$, so the numerator has the form $a^2 - b^2$ (a difference of squares).

Use the identity $a^2 - b^2 = (a - b)(a + b)$ with $a = 2z$ and $b = 5$.

So

Substitute the factored form of the numerator into the original expression:

Because $z \neq -\tfrac{5}{2}$, the denominator $2z + 5$ is never zero, so the common factor $2z + 5$ in the numerator and denominator can be canceled, leaving only one linear factor in the numerator.

After canceling the common factor $(2z + 5)$, the expression simplifies to

This is equivalent to the original expression for all $z \neq -\tfrac{5}{2}$, and it corresponds to answer choice D.