Which expression can be factorised using \(A^2-B^2=(A-B)(A+B)\) without regrouping its terms?

Correct answer: \(p^2-10p+25-49q^2\)

The correct answer is A. Without regrouping, the first three terms of A are already a perfect square: \(p^2-10p+25=(p-5)^2\). The last term is \(49q^2=(7q)^2\), and the sign between these two square expressions is minus. Thus A becomes \((p-5)^2-(7q)^2\), which factors directly as \((p-5-7q)(p-5+7q)\). Option B has a plus sign between the square expressions, so it is a sum of squares, not the required difference. Option C has \(-49q\), which is not the square \((7q)^2\), and its middle sign also changes the perfect square. Option D has \(p^2-25ba\) but the remaining terms do not form one square without regrouping, so it does not fit the requested direct pattern. First identify a complete square and then check the separating sign. Memory cue: complete square minus complete square gives two conjugate factors.