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Factor $4z^2-9\text{.}$

Solution

The “ $4z^2$ ” at the beginning here is a perfect square. It is the same as $(2z)^2\text{.}$ So we can take $A=2z\text{.}$ The “ $9$ ” being subtracted is also a perfect square, so we can take $B=3\text{:}$

$\begin{aligned} 4z^2-9\amp=(A-B)(A+B)\\ \amp=(2z-3)(2z+3) \end{aligned}$