# hw9 - 2 x 1 x 1 x = = x 4 x 1 x 3 x 1 x 4 x 3 1 x 3 x 2 Now...

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MATH 3360 HOMEWORK SOLUTION 9 (Some computational problems are omitted.) p251, E10 Because 0 , 1 , 2 ,...,p - 1 are exactly all roots of this polynomial and Z/pZ is a ﬁeld. p251, E12 You have to show your work, i.e., prove that the polynomials you ﬁnd are irreducible and others are not, using whatever theorem you need. p307, E9(i) Don’t forget constants: 1 and 2 are also units. p309, E2(ii) First use Euclid’s Algorithm to ﬁnd the gcd of x 4 + x + 1 and x 4 + x 3 + 1: x 4 + x 3 + 1 = ( x 4 + x + 1) + ( x 3 + x ) x 4 + x + 1 = ( x 3 + x ) x + ( x 2 + x + 1) x 3 + x = ( x 2 + x + 1)( x + 1) + ( x + 1) x 2 + x + 1 = ( x + 1) x + 1 So the gcd is 1 and for Bezout: ( * ) : 1 = ( x
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Unformatted text preview: 2 + x + 1) + ( x + 1) x = ... = ( x 4 + x + 1)( x 3 + x + 1) + ( x 4 + x 3 + 1)( x 3 + x 2 ) Now to solve f ( x ), assume that: ‰ f ( x ) = r ( x )( x 4 + x + 1) + x 2 f ( x ) = s ( x )( x 4 + x 3 + 1) + x We get r ( x )( x 4 + x + 1) + s ( x )( x 4 + x 3 + 1) = x 2 + x . Using ( * ) we know that r ( x ) = ( x 2 + x )( x 3 + x + 1) and s ( x ) = ( x 2 + x )( x 3 + x 2 ). Finally f ( x ) = r ( x )( x 4 + x +1)+ x 2 = ( x 2 + x )( x 3 + x +1)( x 4 + x +1)+ x 2 = x 9 + x 8 + x 7 + x 6 + x 5 + x 3 + x . 1...
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