What is the factorisation of ( 121-4y^2 )?

Correct answer: \((11+2y)(11-2y)\)

The expression is a difference of two squares: \(121=11^2\) and \(4y^2=(2y)^2\). Using \(a^2-b^2=(a+b)(a-b)\), we get \(121-4y^2=(11+2y)(11-2y)\). The option \((11-2y)^2\) would also produce a middle term \(-44y\), so it is not correct. Exam tip: identify the square roots of both terms before applying the difference-of-squares identity.