How do you factor $3 x {(x - 2 y)}^{2} - {(x - 2 y)}^{3}$ $3x(x - 2y)^2- (x - 2y)^3$ ?

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Answer 1 · 2015-10-09T10:42:58

$2 (x + y) {(x - 2 y)}^{2}$ $2(x+y)(x-2y)^2$

Both terms are multiple of ${(x - 2 y)}^{2}$ $(x-2y)^2$ , so we can collect it:

$3 x {(x - 2 y)}^{2} - {(x - 2 y)}^{3} = {(x - 2 y)}^{2} [3 x - (x - 2 y)]$ $3x(x-2y)^2 - (x-2y)^3 = (x-2y)^2 [3x-(x-2y)]$

Simplifying the square brackets gives

$3 x - (x - 2 y) = 3 x - x + 2 y = 2 x + 2 y = 2 (x + y)$ $3x-(x-2y) = 3x-x+2y = 2x+2y=2(x+y)$

So, the whole expression equals $2 (x + y) {(x - 2 y)}^{2}$ $2(x+y)(x-2y)^2$

How do you factor 3x(x - 2y)^2- (x - 2y)^33x(x−2y)2−(x−2y)33x(x - 2y)^2- (x - 2y)^3?