# What is the final result of (x^2-4x+4)^2 - (x-2)^2 ?

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We recognize a difference of two squares and we'll apply the formula:

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

We'll put a= x^2-4x+4 and b = x-2

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

We'll combine like terms inside brackets:

(x^2-4x+4)-(x-2)^2 = (x^2 - 5x + 6)( x^2-3x + 2)

The roots of the first factor, x^2 - 5x + 6, are:

x1 = 2 and x2 = 3

x^2 - 5x + 6 = (x-2)(x-3)

The roots of the second factor, x^2-3x + 2, are:

x1 = 1 and x2 = 2

x^2-3x + 2 = (x-1)(x-2)

(x^2-4x+4)^2 - (x-2)^2 = (x-2)(x-3)(x-1)(x-2)

**The final result of the difference of 2 squares is: (x^2-4x+4)^2 - (x-2)^2 = (x-3)*(x-1)*(x-2)^2.**