Calculate (x^2+1)-(x^2-4x+1)^2

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We need to find (x^2 + 1) - (x^2 - 4x + 1)^2

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

=> x^2 + 1 - [(x^2 - 4x) + 1]^2

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

=> x^2 + 1 - x^4 - 16x^2 + 8x^3 - 1 - 2x^2 + 8x

=> -x^4 + 8x^3 -17x^2 + 8x

The required value is -x^4 + 8x^3 -17x^2 + 8x

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since it is a difference of squares, we'll apply the formula:

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

We'll put a= x^2+1 and b = x^2-4x+1

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

We'll combine like terms inside brackets:

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

(x^2+1)-(x^2-4x+1)^2 = 8x(x^2 - 2x + 1)

We notice that inside brackets we have a perfect square:

(x^2+1)-(x^2-4x+1)^2 = 8x(x - 1)^2

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