Which is the correct factor form 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 \(a^2-b^2=(a+b)(a-b)\), with \(a=11\) and \(b=2y\), we get \(121-4y^2=(11+2y)(11-2y)\). Option A is the square of a binomial, so it is not the factorisation of a difference of squares. Exam tip: Rewrite both terms as perfect squares before selecting an identity.