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Mathematics 22 Online
OpenStudy (anonymous):

Factor each of the the following difference of the squares 1. x^2-9 2. 36x^2-81

OpenStudy (anonymous):

\[(x-3)(x+3)\]because \[a^2-b^2=(a+b)(a-b)\]

OpenStudy (anonymous):

what about for 36x^2-81

OpenStudy (anonymous):

same idea. see if you see that both \[36x^2\] and \[81\] are "perfect squares" in the form \[a^2-b^2\]

OpenStudy (anonymous):

so would it be (x+6)(x-6)- (9 x 9)?

OpenStudy (anonymous):

It would be (6x-9)(6x+9)

OpenStudy (anonymous):

thank you!

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