354 CHAPTER 3 Polynomial and Rational Functions Using the result of the synthetic division, ƒ1x2 can be written in the following factored form. ƒ1x2 = 3x - 11 - i243x - 11 + i241x2 - 5x + 62 ƒ1x2 = 3x - 11 - i243x - 11 + i241x - 221x - 32 The remaining zeros are 2 and 3. The four zeros are 1 - i, 1 + i, 2, and 3. S Now Try Exercise 35. Descartes’ Rule of Signs The following rule helps to determine the number of positive and negative real zeros of a polynomial function. A variation in sign is a change from positive to negative or from negative to positive in successive terms of the polynomial when they are written in order of descending powers of the variable. Missing terms (those with 0 coefficients) are counted as no change in sign and can be ignored. Descartes’ Rule of Signs Let ƒ1x2 define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. (a) The number of positive real zeros of ƒ either equals the number of variations in sign occurring in the coefficients of ƒ1x2, or is less than the number of variations by a positive even integer. (b) The number of negative real zeros of ƒ either equals the number of variations in sign occurring in the coefficients of ƒ1-x2, or is less than the number of variations by a positive even integer. EXAMPLE 7 Applying Descartes’ Rule of Signs Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of ƒ1x2 = x4 - 6x3 + 8x2 + 2x - 1. SOLUTION We first consider the possible number of positive zeros by observing that ƒ1x2 has three variations in signs. ƒ1x2 = +x4 - 6x3 + 8x2 + 2x - 1 1 2 3 Thus, ƒ1x2 has either three or one (because 3 - 2 = 1) positive real zeros. For negative zeros, consider the variations in signs for ƒ1-x2. ƒ1-x2 = 1-x24 - 61-x23 + 81-x22 + 21-x2 - 1 ƒ1-x2 = x4 + 6x3 + 8x2 - 2x - 1 1 There is only one variation in sign, so ƒ1x2 has exactly one negative real zero. NOTE If we had been unable to factor x2 - 5x + 6 into linear factors in Example 6, we would have used the quadratic formula to solve the equation x2 - 5x + 6 = 0 to find the remaining two zeros of the function.
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