How do you factor #9(x + 2y + z)^2 - 16(x - 2y + z)^2#?

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Answer 1 · 2015-07-04T03:49:53

This is a difference of squares.

Difference of squares: #u^2 - v^2 = (u+v)(u-v)#

#9(x + 2y + z)^2 = (3(x+2y+z))^2# That will be our #u^2#

#16(x - 2y + z)^2 = (4(x - 2y + z))^2#. That will be our #v^2#

#9(x + 2y + z)^2 - 16(x - 2y + z)^2 = (3(x+2y+z))^2 - (4(x - 2y + z))^2#

#=[3(x+2y+z) + 4(x - 2y + z)] [3(x+2y+z) - 4(x - 2y + z)]#

Each of these factors can be simplified, to get:

#(7x-2y+7z)(-x+14y-z)#