Factorise \(px+qx+py+qy\) by grouping.

Correct answer: \((p+q)(x+y)\)

The direct answer is A: \((p+q)(x+y)\). Group the terms containing \(x\) and the terms containing \(y\): \(px+qx+py+qy=x(p+q)+y(p+q)\). The same binomial \((p+q)\) is common in both terms. Taking it outside gives \((p+q)(x+y)\). Expanding confirms the result: \((p+q)(x+y)=px+py+qx+qy\), which is the same set of terms as the original expression, only in a different order. B expands to \(pq+py+qx+xy\), so it introduces terms that were not given. C expands to \(pq+px+qy+xy\), also incorrect. D gives only two terms, \(pqx+pqy\), and omits the required separate coefficients. Grouping is successful when the same factor appears after factoring each group.