How do you factor $9x^2y^2-z^2$ ?

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Answer 1 · 2015-07-06T18:39:58

This is a difference of squares:

$9x^2y^2-z^2 = (3xy)^2 - z^2 = (3xy-z)(3xy+z)$

The difference of squares identity is:

$a^2-b^2 = (a-b)(a+b)$

In our example, let $a=3xy$ and $b=z$ ...

$9x^2y^2-z^2$

$=(3xy)^2-z^2$

$=a^2-b^2$

$=(a-b)(a+b)$

$=(3xy-z)(3xy+z)$

How do you factor 9x^2y^2-z^2 9x^2y^2-z^2 ?