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If \(p(x)=x^3-9x\), how many distinct real zeroes does it have?

Answer and explanation

Correct answer: 3

Factoring gives \(p(x)=x(x^2-9)=x(x-3)(x+3)\). Thus, the zeroes are \(x=0,3,-3\), which are three distinct real numbers. Therefore, the correct answer is 3. Exam tip: when counting distinct zeroes, count a repeated zero only once.

Related tags

PolynomialsDistinct-ZeroesCubic-PolynomialFactorisationReal-ZeroesPolynomials In One VariableMathematicsClass 10 Mcq

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