Quiz13 - Quiz 13 Solution Fall 2010 October 6 This problem...

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Quiz 13 Solution Fall 2010 October 6 This problem is from Lesson 7 material. Find the zeros (roots) of 24 64 38 34 23 3 ) ( 2 3 4 5 - - + + - = x x x x x x f . Also, give a factorization for the polynomial. Factors of 24 - : 1 ± , 2 ± , 3 ± , 4 ± , 6 ± , 8 ± , 12 ± , 24 ± Factors of 3: 1, 3 Possible rational zeros (roots): 1 ± , 2 ± , 3 ± , 4 ± , 6 ± , 8 ± , 12 ± , 24 ± , 3 1 ± , 3 2 ± , 3 4 ± , 3 8 ± Trying 1: 36 12 52 14 20 3 12 52 14 20 3 24 64 38 34 23 3 24 64 38 34 23 3 2 3 4 5 - - - - - - - - - - + + - x x x x x of ts Coefficien 1 Thus, 1 0 36 ) 1 ( - - = x f is not a factor of f and 1 is not a zero (root) of f . Also, 1 is not an upper bound for the positive zeros (roots). Trying 1 - : 18 42 22 60 26 3 42 22 60 26 3 24 64 38 34 23 3 24 64 38 34 23 3 2 3 4 5 - - - - - - - - - - + + - x x x x x of ts Coefficien 1 - Thus, 1 0 18 ) 1 ( + = - x f is not a factor of f and 1 - is not a zero (root) of f . Also, 1 - is not a lower bound for the negative zeros (roots).
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Trying 2: 0 12 38 0 17 3 24 76 0 34 6 24 64 38 34 23 3 24 64 38 34 23 3 2 3 4 5 - - - - - - - + + - x x x x x of ts Coefficien 2 Thus, 2 0 ) 2 ( - = x f
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Quiz13 - Quiz 13 Solution Fall 2010 October 6 This problem...

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