Sunday, February 19, 2012

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

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.

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