Factorise \(3a+3b+xa+xb\).

Correct answer: \((3+x)(a+b)\)

The direct answer is A: \((3+x)(a+b)\). Arrange the four terms in two useful groups: \(3a+3b=3(a+b)\) and \(xa+xb=x(a+b)\). Thus the expression becomes \(3(a+b)+x(a+b)\). The common factor is the whole bracket \((a+b)\), so taking it out gives \((3+x)(a+b)\). Expanding A confirms this: \(3a+3b+xa+xb\). Option B would expand to \(3a-3b+xa-xb\), changing both signs involving \(b\). Option C would give \(3a+3b-xa-xb\), changing the signs of the \(x\)-terms. Option D gives \(3ax+3bx\), which has products not present in the original. Common-confusion warning: do not multiply the outside coefficients together; they are being added after the shared bracket is identified.