If a visual model subtracts two \(4y\) rectangles from a \(y^2\) square and adds a \(16\) square, what is the identity?

Correct answer: \((y-4)^2=y^2-8y+16\)

Direct answer: option A, \((y-4)^2=y^2-8y+16\). Start with the square of side \(y\), whose area is \(y^2\). Removing two rectangles of area \(4y\) removes \(4y+4y=8y\). Adding the small square of side 4 contributes \(4^2=16\). Thus the resulting expression is \(y^2-8y+16\). Using \((a-b)^2=a^2-2ab+b^2\), take \(a=y\) and \(b=4\), giving \((y-4)^2=y^2-2(y)(4)+4^2=y^2-8y+16\). Option B has a positive middle term and represents addition. Option C is the difference-of-squares identity and has no \(-8y\) term. Option D incorrectly equates a difference of squares with a perfect-square expansion. Memory cue: subtracting two equal rectangles means the middle term is \(-2ab\), not zero.