How do you factor #x^2-4y^2-4x+4# by grouping?

1 Answer
Feb 1, 2017

Answer:

#x^2-4y^2-4x+4 = (x-2y-2)(x+2y-2)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

By rearranging the given quadratic, we can recognise it as a difference of squares and hence factor it:

#x^2-4y^2-4x+4 = (x^2-4x+4)-4y^2#

#color(white)(x^2-4y^2-4x+4) = (x-2)^2-(2y)^2#

#color(white)(x^2-4y^2-4x+4) = ((x-2)-2y)((x-2)+2y)#

#color(white)(x^2-4y^2-4x+4) = (x-2y-2)(x+2y-2)#