Descartes rule of signs Y= x^3 - x-1 How many zeros does this function have? I t

Question

Descartes rule of signs

Y= x^3 - x-1

How many zeros does this function have? I thought using "Descartes rule of signs" it would be one real, but why not follow that forumula to also get one neg zero?

Doesn't it work the same way? Plugging in -x for x would still make for one sign change....

I see the graph isn't that way but why is this not working out using the rule of signs? Does it have to do with the fact that complex zeros have to come in pairs and if there was one neg zero it would only leave a count for one complex?

Thanks

Explanation / Answer

Descartes rule states that the number of sign changes in f(x) are the MAXIMUM NUMBER of positive roots.. and the number of sign changes of f(-x) are the MAXIMUM NUMBER of negative roots.

Notice that Descartes rule tells you the maximum positive/Negative roots that the function can have and not the absolute numbers.

So here, the function can have either 1 positive real root and 2 imaginary roots (since imaginary roots always occur in conjugate pairs) or 1 negative real root and 2 imaginary roots.

1 positive real root and 1 negative real root is not possible because that would leave just 1 count for imaginary root which is not possible (As you have rightly stated)

Hope this makes it clear!

Your follow up questions are most welcome!

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