Problem

Polynomial and Rational Functions Descartes' Rule of Signs Use Descartes' Rule of Signs to determine the possible numbers of positive and negative real zeros. \[ f(x)=4 x^{4}-5 x^{3}-7 x^{2}+6 x+7 \] If there is more than one possibility, separate them with commas. (a) Possible number(s) of positive real zeros: (b) Possible number(s) of negative real zeros:

Solution

Step 1 :Apply Descartes' Rule of Signs to the given polynomial \(f(x)=4 x^{4}-5 x^{3}-7 x^{2}+6 x+7\) to find the possible number of positive real zeros.

Step 2 :Count the number of sign changes in the polynomial. The coefficients of the polynomial go from positive to negative twice (from \(4x^4\) to \(-5x^3\) and from \(-7x^2\) to \(6x\)), so there are 2 sign changes.

Step 3 :According to Descartes' Rule of Signs, the number of positive real zeros is either equal to the number of sign changes, or less than that by a multiple of 2. So, the possible number of positive real zeros is \(\boxed{2}\) or \(\boxed{0}\).

Step 4 :Apply Descartes' Rule of Signs to the polynomial \(f(-x)=4 x^{4}+5 x^{3}-7 x^{2}-6 x+7\) to find the possible number of negative real zeros.

Step 5 :Count the number of sign changes in the polynomial. The coefficients of the polynomial go from positive to negative twice (from \(5x^3\) to \(-7x^2\) and from \(-6x\) to \(7\)), so there are 2 sign changes.

Step 6 :According to Descartes' Rule of Signs, the number of negative real zeros is either equal to the number of sign changes, or less than that by a multiple of 2. So, the possible number of negative real zeros is \(\boxed{2}\) or \(\boxed{0}\).

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