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Reema factorised \(x^2-9\) as \((x-3)^2\). What is the correct factorisation after fixing her error?
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Answer and explanation
Correct answer: \((x-3)(x+3)\)
\(x^2-9=x^2-3^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-3)(x+3)\). In contrast, \((x-3)^2=x^2-6x+9\). Exam tip: check the middle term by expanding.
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What is the correct answer to this question?
\((x-3)(x+3)\)
Why is this the correct answer?
\(x^2-9=x^2-3^2\) is a difference of squares. Using \(a^2-b^2=(a-b)(a+b)\), it becomes \((x-3)(x+3)\). In contrast, \((x-3)^2=x^2-6x+9\). Exam tip: check the middle term by expanding.
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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