Which is the correct factorisation of \(x^2-7x+10\)?

Correct answer: \((x-2)(x-5)\)

The direct answer is B: \((x-2)(x-5)\). We need two numbers whose product is the constant term 10 and whose sum is the coefficient of \(x\), namely \(-7\). The numbers \(-2\) and \(-5\) satisfy both conditions. Thus \(x^2-7x+10=(x-2)(x-5)\). On expansion, \(x^2-5x-2x+10=x^2-7x+10\). Option A gives factors 1 and 10, whose sum is 11, so it produces \(-11x\), not \(-7x\). Option C gives a positive middle term \(+7x\), so its sign is wrong. Option D uses 3 and 4; it gives \(-7x\) but its product is 12, not 10. B is the only option satisfying both tests. Memory cue: the constant term checks the product, while the middle coefficient checks the signed sum.