Which expansion is correct for the area of a square with side \((x-7)\)?

Correct answer: \(x^2-14x+49\)

Correct answer: B, \(x^2-14x+49\). The square’s area is \((x-7)^2\). Apply \((p-q)^2=p^2-2pq+q^2\) with \(p=x\) and \(q=7\): \(x^2-2(x)(7)+7^2=x^2-14x+49\). In a visual model, the large \(x\times x\) square contributes \(x^2\), two strips of area \(7x\) are removed, and the \(7\times7\) corner is added back, giving \(-14x+49\). Option A subtracts only one strip. Option C is \((x+7)(x-7)\), a difference of squares, not \((x-7)^2\). Option D has the sign for \((x+7)^2\). Memory cue: squaring a difference creates a negative double-product term, not merely \(-7x\).