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)#