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

1 Answer
smendyka
Jun 17, 2017

See a solution process below:

Explanation:

This is a special form of the quadratic where:

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

Substituting #y# for #a# and substituting #2x# for #b# gives:

#(y - 2x)^2 = y^2 - 2y2x + (2x)^2#

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

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