Which is the correct factorisation of \(3y^2 − 14y − 5\)?

Correct answer: (3y + 1)(y − 5)

The direct answer is A. Expand each factor in A: \((3y+1)(y−5)=3y^2−15y+y−5=3y^2−14y−5\), exactly the given expression. Another way is to use the ac method: \(3×(−5)=−15\), and the numbers \(−15\) and \(1\) have sum \(−14\). Split the middle term as \(−15y+y\), then group: \(3y(y−5)+1(y−5)=(3y+1)(y−5)\). B expands to \(3y^2+15y−y−5=3y^2+14y−5\), so the middle sign is wrong. C has leading coefficient \(3\) but its constant term is \(−15\), not \(−5\). D expands to \(3y^2−2y−5\), whose middle coefficient is different. Always verify a proposed factorisation by multiplying; this is especially useful when signs are mixed. Memory cue: the product of the constants must equal \(−5\), while the middle terms must combine to \(−14y\).