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Chapter 2: Problem 34

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. $$f(x)=x^{3}+7 x^{2}+x+7$$

Short Answer

Expert verified
Based on Descartes's Rule, there are no positive real roots and one negative real root in the polynomial \(f(x) = x^3 + 7x^2 + x + 7\).

Step by step solution

01

Identifying the Polynomial and its Coefficients

The given polynomial is \(f(x) = x^3 + 7x^2 + x + 7\). The coefficients of the terms are 1, 7, 1, and 7 respectively. The order of these coefficients matters for applying Descartes's Rule, and we shall go from highest degree term to lowest degree.
02

Counting Sign Changes for Positive Roots

The signs of the coefficients are +, +, +, +. As we can see, there are no changes in the sign while we move from the highest degree coefficient to the lowest. Hence, according to Descartes's Rule, there are no positive real roots.
03

Coefficient Changes for Negative Root(s)

Descartes's Rule of signs requires a reflection over y-axis for counting negative real zeros. This is equivalent to substituting \(x\) with \(-x\) in the function. Doing so we get \(f(-x) = -x^3 - 7x^2 - x + 7\). The coefficients are now -, -, -, +.
04

Counting Sign Changes in Coefficients List for Negative Roots

The changes of signs now from high degree to the lowest are -, -, -, +. Therefore, we have one change in sign here. Hence, there is one negative real root according to Descartes's Rule.

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