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Chapter 2: Problem 38

Use Descartes's Rule of Signs to determine the possible number of positive and negative real zeros for each given function. $$f(x)=4 x^{4}-x^{3}+5 x^{2}-2 x-6$$

Short Answer

Expert verified
The possible number of positive real zeros for the function is 2 or 0, and the possible number of negative real zeros is 1.

Step by step solution

01

Finding the Possible Number of Positive Real Zeros

Look at the coefficients of the polynomial in its standard form: \(4, -1, 5, -2, -6\). Find the number of sign changes. The coefficients go from positive (+4) to negative (-1), positive (+5) to negative (-2), and then, finally, negative (-2) to negative (-6). Hence, there are 2 changes of signs.
02

Determine the Possible Number of Positive Real Zeros

According to Descartes's rule, the number of positive real zeros is either equal to the number of sign changes (2) or that number minus an even integer. Therefore, the possible number of positive real zeros is 2 or 0.
03

Finding the Possible Number of Negative Real Zeros

Replace \(x\) with \(-x\) and write the polynomial: \(4 (-x)^{4}-(-x)^{3}+5 (-x)^{2}-(-2 x)-6\). Simplify this to: \(4x^4+x^3+5x^2+2x-6\). Find the number of sign changes. The coefficients go from positive (+4) to positive (+1), positive (+5) to positive (+2), and then positive (+2) to negative (-6). Hence, there's 1 change of sign.
04

Determine the Possible Number of Negative Real Zeros

According to Descartes's rule, the number of negative real zeros is either equal to the number of sign changes (1) or that number minus an even integer. Since we cannot subtract an even number from 1 and get a non-negative result, the possible number of negative real zeros is 1.

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