Question 870177
First factor {{{4x^2 - 4x + 1}}} to get



{{{4x^2 - 4x + 1}}}



{{{4x^2 - 2x - 2x + 1}}} Note: a*c = 4*1 = 4, the two numbers -2 and -2 multiply to 4 and add to -4, so that's why I broke up -4x into -2x-2x



{{{(4x^2 - 2x) + (-2x + 1)}}}



{{{2x(2x - 1) + (-2x + 1)}}}



{{{2x(2x - 1) - 1(2x - 1)}}}



{{{(2x - 1)(2x - 1)}}}



{{{(2x - 1)^2}}}



Note: {{{4x^2 - 4x + 1}}} is a perfect square trinomial, so you can use the special factoring formula for that.



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So 



{{{4x^2 - 4x + 1 - y^2}}} 



turns into 



{{{(2x - 1)^2-y^2}}}



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Looking at {{{(2x - 1)^2-y^2}}} we have a difference of squares. So use the difference of squares factoring formula to go from {{{(2x - 1)^2-y^2}}} to {{{(2x - 1-y)(2x-1+y)}}}. You can rearrange things to get {{{(2x-y-1)(2x+y-1)}}}



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The final answer is {{{(2x-y-1)(2x+y-1)}}}