Which is the correct factorisation of (2x^2+7xy+3y^2)?

Correct answer: ( (2x+y)(x+3y) )

The direct answer is option B: \((2x+y)(x+3y)\). To factor, multiply the first terms to get \(2x^2\) and the last terms to get \(3y^2\). The middle products are \(6xy\) and \(xy\), whose sum is \(7xy\). Thus \(2x^2+6xy+xy+3y^2=2x^2+7xy+3y^2\). Option A gives \(2x^2+5xy+3y^2\), so its middle term is wrong. Option B gives exactly the expression, so it is correct. Option C gives \(2x^2+7xy+6y^2\), so its last term is wrong. Option D gives \(2x^2-7xy+3y^2\), so the middle sign is wrong. Memory cue: for \(ac=6\) and sum \(7\), use \(1\) and \(6\).