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Re: (x^2+9)(x+9)(x-5)=0 [#permalink]
The basic idea in this questions is that A x B =0 only when A=0 or B=0 or both=0

Similarly, in the given question, one of the terms has to be 0,
thus we have a total of 3 possible equations.

\(x^2+9=0\) implying that x is an imaginary number since square of any real number is always positive
\(x+9=0\) implying that x can be -9
\(x-5=0\) implying that x can be 5

Both the real possible values of x are lesser than part B of the question, hence Option B is correct because 7 will always be greater than -9 or 5
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