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Chapter 2: Problem 27

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=x^{3}+3 x^{2}-6 x-8$$

Short Answer

Expert verified
The number of possible positive real zeros can be either 2 or 0, and the number of possible negative real zeros can also be either 2 or 0.

Step by step solution

01

Determine the Number of Sign Changes in the Polynomial Coefficients

First, list out the coefficients of the polynomial \(P(x)=x^{3}+3 x^{2}-6 x-8\), which are 1, 3, -6, and -8. There are two sign changes, (from 3 to -6 and from -6 to -8). This means there could be 2 or \(2-2=0\) positive real zeros.
02

Find the Number of Sign Changes for the Polynomial with \(x\) Replaced by \(-x\)

Replace \(x\) with \(-x\) and simplify the resulting polynomial. It becomes \(P(-x)=(-x)^3+3(-x)^2-6(-x)-8=-x^{3}+3 x^{2}+6 x-8\). The coefficients now are -1, 3, 6, and -8. There are two sign changes (from -1 to 3 and from 6 to -8). This means there could be 2 or \(2-2=0\) negative real zeros.

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