Write (4r^2-81) in factors.
Answer and explanation
Correct answer: \((2r-9)(2r+9)\)
This is a difference of two perfect squares because \(4r^2=(2r)^2\) and \(81=9^2\). Applying \(a^2-b^2=(a-b)(a+b)\) with \(a=2r\) and \(b=9\) gives \(4r^2-81=(2r-9)(2r+9)\). The close distractor \((2r-9)^2\) expands to \(4r^2-36r+81\), which has a middle term and a positive constant term. Exam tip: before factorising, check whether both terms are perfect squares separated by a minus sign.
Frequently asked questions
What is the correct answer to this question?
\((2r-9)(2r+9)\)
Why is this the correct answer?
This is a difference of two perfect squares because \(4r^2=(2r)^2\) and \(81=9^2\). Applying \(a^2-b^2=(a-b)(a+b)\) with \(a=2r\) and \(b=9\) gives \(4r^2-81=(2r-9)(2r+9)\). The close distractor \((2r-9)^2\) expands to \(4r^2-36r+81\), which has a middle term and a positive constant term. Exam tip: before factorising, check whether both terms are perfect squares separated by a minus sign.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Exploring Algebraic Identities. Topic: Factorisation.
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