Which is the correct factorisation of \(2a^2-a-15\)?

Correct answer: \((2a+5)(a-3)\)

The direct answer is A: \((2a+5)(a-3)\). Multiply the leading coefficient and constant: \(2\times(-15)=-30\). We need two numbers with product \(-30\) and sum \(-1\): they are \(5\) and \(-6\). Split the middle term: \(2a^2-a-15=2a^2+5a-6a-15\). Grouping gives \(a(2a+5)-3(2a+5)=(2a+5)(a-3)\). Checking option A directly gives \(2a^2-6a+5a-15=2a^2-a-15\). B gives a positive constant, not the required negative one. C gives \(2a^2+7a-15\), so its middle term is wrong. D gives \(2a^2-31a+15\). A negative constant usually means the two constant signs in the factors are opposite, but the middle coefficient must still be checked.