Explanation Theorem used: Rational Root Theorem: Any rational root r r of a polynomial equation f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 0 = 0 is an integer factor of a 0 divided by an integer factor of a n . Therefore, r r = integer factor of a 0 integer factor of a n . Definition used: Descartes’ Rule of Signs: The number of positive of a polynomial equation f ( x ) = 0 cannot be greater than the number of changes in sign in f ( x ) in moving from one term to the next in f ( x ) . The number of negative cannot be greater than the number of sign changes in f ( − x ) . Calculation: It is given that the polynomial equation is D 5 + D 4 − 9 D 3 − 5 D 2 + 16 D + 12 = 0 . Here, f ( D ) = D 5 + D 4 − 9 D 3 − 5 D 2 + 16 D + 12 and the degree is 50. Move from left to right, obtain the number of sign changes. There are 2 sign changes; hence the number of positive roots is 2. Change the function f ( D ) to f ( − D ) as follows. f ( − D ) = ( − D ) 5 + ( − D ) 4 − 9 ( − D ) 3 − 5 ( − D ) 2 + 16 ( − D ) + 12 = − D 5 + D 4 + 9 D 3 − 5 D 2 − 16 D + 12 As there are 3 sign changes, the number of negative roots is 3. Hence, the maximum possible numbers of positive and negative roots of D 5 + D 4 − 9 D 3 − 5 D 2 + 16 D + 12 = 0 are 2 and 3, respectively. Thus, the possible rational roots by Rational Root Theorem are, r r = integer factor of 12 integer factor of 1 = 1 , 2 , 3 , 4 , 6 , 12 1 = ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 12 Begin with − 1 , perform synthetic division as follows. 1 1 − 9 − 5 16 12 − 1 0 9 − 4 − 12 1 0 − 9 4 12 0 [ − 1 ] The remainder is 0

11th Edition

ISBN: 9780134508290

Author: Evans

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 15.3, Problem 17E

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Chapter 15.3, Problem 16E

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Let Χ be a real-valued character (mod k). Let k S = Σnx(n). n=1 If (a, k) = 1, ax(a)S = S (mod k). (iii) Write k = 2ºq where q is odd. Show that there is an integer a with (a, k) = 1 such that a = 3 (mod 2ª) and a = 2 (mod q). Deduce that 12S = 0 (mod k).

Chapter 15 Solutions

EBK BASIC TECHNICAL MATHEMATICS

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