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Using Descartes' Rule of Signs, we can tell that the polynomial P(x) = x5 − 2x4 + 5x3 − x2 + 4x − 4 has, from smallest to largest, 1 , 2 , or 3 positive real zeros and 4 negative real zeros.

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Answer:

In the question,

According to Descartes' Rule of signs change we can say that the number of sign changes of the co-efficient of the polynomial is the number of positive zeroes of the polynomial.

And,

On putting x = -x,

The number of sign changes a polynomial obtained has, is equal to the number of negative zeroes.

So,

In the polynomial,

[tex]P(x)=x^{5}-2x^{4}+5x^{3}-x^{2}+4x-4[/tex]

So, we can see that the number of sign changes are from 1 to -2, -2 to 5, 5 to -1, -1 to 4, 4 to -4.

So, there are 5 number of co-efficient sign changes taking place.

Therefore, there are 5 positive zeroes.

Now,

at x = -x,

[tex]P(-x)=(-x)^{5}-2(-x)^{4}+5(-x)^{3}-(-x)^{2}+4(-x)-4\\P(-x)=-x^{5}-2x^{4}-5x^{3}-x^{2}-4x-4[/tex]

Here, we can see that there are no sign changes in the polynomial.

Therefore, there are 0 negative zeroes.

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