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:

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

How do you factor 9(x + 2y + z)^2 - 16(x - 2y + z)^29(x + 2y + z)^2 - 16(x - 2y + z)^2?