How do you factor #81p^2 - 144y^2#?

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Answer 1 · 2016-09-27T19:24:51

#(9p+12y)(9p-12y)#

If you look carefully at the two terms in the polynomial, you'll quickly figure out that they are square numbers.

Thus, you can write them like this:

#81p^2-144y^2# is the same as #(9p)^2-(12y)^2#

This actually looks like #a^2-b^2#, where #a=9p# and #b=12y#

Recall the identity, #a^2-b^2=(a+b)(a-b)#

Now you just have to put it in that form!

#(9p)^2-(12y)^2=(9p+12y)(9p-12y)#