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Chapter 2: Problem 35

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. $$f(x)=5 x^{3}-3 x^{2}+3 x-1$$

Short Answer

Expert verified
The possible number of positive real zeros for the function \(f(x) = 5x^3 - 3x^2 + 3x - 1\) is 3 or 1 and the possible number of negative real zeros is 1.

Step by step solution

01

Determine possible positive real zeros

Write the function in standard form (decreasing powers of x), which in this case is \(f(x)=5x^3 - 3x^2 + 3x - 1\). Count the number of sign changes. The coefficient starts positive (5), then becomes negative (-3), then positive (3) and ends as negative (-1). Therefore there are three sign changes. According to Descartes' Rule, the number of positive real zeros is either the same as the number of sign changes or it is less than this by a multiple of two. So, the possible number of positive real zeros is 3 or 3-2=1.
02

Determine possible negative real zeros

To find the possible negative real zeros, replace each \(x\) in the polynomial by \(-x\), resulting in the equation \(f(-x)=5(-x)^3 - 3(-x)^2 + 3(-x) - 1 = -5x^3 - 3x^2 -3x - 1\). There is only one sign change (from -5 to -3), so the possible number of negative real zeros is 1 or 1-2=-1. But we can't have a negative number of zeros, so we only consider 1.
03

Summary

Combine the results from step 1 and step 2 to get the final answer. The possible number of positive real zeros is 3 or 1, and the possible number of negative real zeros is 1.

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