# Adv Alegbra HW Solutions 113 - 7 ² 1 3 ² 7 3 Solution...

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113 Since n is odd, ( x ) n 1 =− x n 1 =− ( x n + 1 ), and one can now cancel the minus sign from each side. (ii) For numbers a and b , prove that a n b n = ( a b )( a ζ b )( a ζ 2 b ) ··· ( a ζ n 1 b ) and, if n is odd, that a n + b n = ( a + b )( a + ζ b )( a + ζ 2 b ) ··· ( a + ζ n 1 b ). Solution. If b = 0, then both sides equal a n ;i f b ±= 0, then set x = a / b in part (ii). 3.86 True or false with reasons. (i) 3 is an algebraic integer. Solution. True. (ii) 13 78 is not a rational root of 1 + 5 x + 6 x 2 . Solution. False. (iii) If f ( x ) = 3 x 4 + ax 3 + bx 2 + cx + 7 with a , b , c Z , then the roots of f ( x ) in Q , if any, lie in 1 , ² 7 , ² 1 3 , ² 7 3 } . Solution. True. (iv) If f ( x ) = 3 x 4 + ax 3 + bx 2 + cx + 7 with a , b , c Q , then the roots of f ( x ) in Q , if any, lie in 1 ,
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Unformatted text preview: 7 , ² 1 3 , ² 7 3 } . Solution. False. (v) 6 x 2 + 10 x + 15 is a primitive polynomial. Solution. True. (vi) Every primitive polynomial in Z [ x ] is irreducible. Solution. False. (vii) Every irreducible polynomial in Z [ x ] is primitive. Solution. False. (viii) Every monic polynomial in Z [ x ] is primitive. Solution. True. (ix) The content of 3 x + 1 5 is 3 5 . Solution. False. (x) The content of 3 x + 6 5 is 3 5 . Solution. True....
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