368 CHAPTER 3 Polynomial and Rational Functions The boundedness theorem shows how the bottom row of a synthetic division is used to place upper and lower bounds on possible real zeros of a polynomial function. Boundedness Theorem Let ƒ1x2 be a polynomial function of degree n Ú 1 with real coefficients and with a positive leading coefficient. Suppose ƒ1x2 is divided synthetically by x - c. (a) If c 70 and all numbers in the bottom row of the synthetic division are nonnegative, then ƒ1x2 has no zero greater than c. (b) If c 60 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then ƒ1x2 has no zero less than c. EXAMPLE 6 Using the BoundednessTheorem Show that the real zeros of ƒ1x2 = 2x4 - 5x3 + 3x + 1 satisfy these conditions. (a) No real zero is greater than 3. (b) No real zero is less than -1. SOLUTION (a) Because ƒ1x2 has real coefficients and the leading coefficient, 2, is positive, use the boundedness theorem. Divide ƒ1x2 synthetically by x - 3. 3)2 -5 0 3 1 6 3 9 36 2 1 3 12 37 All are nonnegative. Here 3 70 and all numbers in the last row of the synthetic division are nonnegative, so ƒ1x2 has no real zero greater than 3. (b) We use the boundedness theorem again and divide ƒ1x2 synthetically by x - 1-12, or x + 1. -1)2 -5 0 3 1 -2 7 -7 4 2 -7 7 -4 5 These numbers alternate in sign. Here -1 60 and the numbers in the last row alternate in sign, so ƒ1x2 has no real zero less than -1. S Now Try Exercises 57 and 59. Proof We outline the proof of part (a). The proof for part (b) is similar. By the division algorithm, if ƒ1x2 is divided by x - c, then for some q1x2 and r, ƒ1x2 = 1x - c2q1x2 + r, where all coefficients of q1x2 are nonnegative, r Ú 0, and c 70. If x 7c, then x - c 70. Because q1x2 70 and r Ú 0, ƒ1x2 = 1x - c2q1x2 + r 70. This means that ƒ1x2 will never be 0 for x 7c.
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