If the final area of a square with side \(q-4\) is needed, which expansion is correct?

Correct answer: \(q^2-8q+16\)

The correct answer is C: \(q^2-8q+16\). A square has area equal to the square of its side, so the required area is \((q-4)^2\). Use the identity \((a-b)^2=a^2-2ab+b^2\), with \(a=q\) and \(b=4\). The first term is \(q^2\). The middle term is \(-2\times q\times4=-8q\), because the two cross-products are both negative. The last term is \(4^2=16\). Hence \((q-4)^2=q^2-8q+16\). Option A misses the factor 2 in the middle term. Option B has the wrong positive sign and corresponds to \((q+4)^2\). Option D has an incorrect middle coefficient and an incorrect constant term. A useful check is to substitute \(q=4\): the original area is zero, and only option C gives \(16-32+16=0\). Remember that a subtraction inside a square gives a negative middle term, not simply \(-q\times4\).