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Chapter 2: Problem 32

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=2 x^{3}+9 x^{2}-2 x-9$$

Short Answer

Expert verified
The given polynomial has 1 positive real zero and no negative real zeros.

Step by step solution

01

Identify sign changes for positive real zeros

Write the polynomial in standard form and count the number of sign changes. The given polynomial is already in standard form \(2 x^{3}+9 x^{2}-2 x-9\). It changes sign once, from \(-2x\) to \(-9\). Therefore, there is either 1 or 1 less an even number (which is -1, but we cannot have negative roots) positive real roots. So, there is 1 positive real root.
02

Identify sign changes for negative real zeros

To find the number of negative real zeros, replace \(x\) with \(-x\) in the polynomial and then count the number of sign changes. This gives us \(2 (-x)^{3}+9 (-x)^{2}-2 (-x)-9 = -2x^3+9x^2+2x-9\). This polynomial does not change its sign. Therefore, there are no negative real zeros.

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