SECTION 4.3 The Real Zeros of a Polynomial Function 223 If the leading coefficient of f is negative, the upper and lower bounds can still be found by first multiplying the polynomial by 1− . Since f x f x 1 , ( ) ( ) ( ) − = − the zeros of f x( ) − are the same as the zeros of f x( ). Proof (Outline) We give only an outline of the proof of the first part of the theorem. Suppose that M is a positive real number, and the third row in the process of synthetic division of the polynomial f by x M− contains only numbers that are positive or 0. Then there are a quotient q and a remainder R for which f x x M q x R ( ) ( ) ( ) = − + where the coefficients of q x( ) are positive or 0 and the remainder R 0 ≥ . Then, for any x M, > we must have x M q x R 0, 0, and 0, ( ) − > > ≥ so that f x 0 ( ) > . That is, there is no zero of f larger than M. The proof of the second part follows similar reasoning. ■ NOTE When finding a lower bound, a 0 can be treated as either positive or negative, but not both. For example, the numbers 3, 0, 5 would be considered to alternate in sign, whereas − 3, 0, 5 would not. j Finding Upper and Lower Bounds of Zeros For the polynomial function f x x x x 2 11 7 6, 3 2 ( ) = + − − use the Bounds on Zeros Theorem to find integer upper and lower bounds to the zeros of f . Solution EXAMPLE 8 NOTE Keep track of any zeros that are found when looking for bounds. j In finding bounds, it is preferable to find the smallest upper bound and largest lower bound. This will require repeated synthetic division until a desired pattern is observed. For simplicity, we consider only potential rational zeros that are integers. If a bound is not found using these values, continue checking positive and/or negative integers until you find both an upper and a lower bound. From Example 4, the potential rational zeros of f are 1, 2, 3, 6, 1 2 , 3 2 ± ± ± ± ± ± . To find an upper bound, start with the smallest positive integer that is a potential rational zero, which is 1. Continue checking 2, 3, and 6 (and then subsequent positive integers), if necessary, until an upper bound is found.To find a lower bound, start with the largest negative integer that is a potential rational zero, which is 1− . Continue checking − − 2, 3, and 6− (and then subsequent negative integers), if necessary, until a lower bound is found. Table 7 summarizes the results of doing repeated synthetic divisions by showing only the third row of each division. For example, the first row of the table shows the result of dividing f x( ) by x 1 − . ) − − 1 2 11 7 6 2 13 6 2 13 6 0 r Coefficients of q x( ) Remainder 1 2 13 6 0 −1 2 9 −16 10 −2 2 7 −21 36 −3 2 5 −22 60 −6 2 −1 −1 0 −7 2 −3 14 −104 Table 7 Synthetic Division Summary Upper bound All nonnegative Lower bound Alternating signs For r 1, = the third row of synthetic division contains only numbers that are positive or 0, so we know there are no real zeros greater than 1. Since the third row of synthetic division for r 7 = − results in alternating positive (or 0) and negative (or 0) values, we know that 7− is a lower bound. There are no real zeros less than 7− . Notice that in looking for bounds, two zeros were discovered. These zeros are 1 and 6− .
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