4.1%20Polynomial%20Arithmetic%20II

4.1%20Polynomial%20Arithmetic%20II - Math 302 Modern...

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Modern Algebra Spring 2009 Homework Solutions 4.1 POLYNOMIAL ARITHMETIC AND THE DIVISION ALGORITHM 5. Find polynomials q ( x ) and r ( x ) such that f ( x ) = g ( x ) q ( x )+ r ( x ) and r ( x ) = 0 or deg r ( x ) < deg g ( x ). (a) f ( x ) = 3 x 4 2 x 3 + 6 x 2 x + 2 and g ( x ) = x 2 + x + 1 in Q [ x ]. x 2 + x + 1 ¢ 3 x 2 5 x + 8 3 x 4 2 x 3 + 6 x 2 x + 2 3 x 4 + 3 x 3 + 3 x 2 5 x 3 + 3 x 2 x + 2 5 x 3 5 x 2 5 x 8 x 2 + 4 x + 2 8 x 2 + 8 x + 8 4 x 6 Then f ( x ) = g ( x )(3 x 2 5 x + 8) + ( 4 x 6) . (b) f ( x ) = x 4 7 x + 1 and g ( x ) = 2 x 2 + 1 in Q [ x ] 2 x 2 + 1 ¢ 1 2 x 2 1 4 x 4 + 0 x 3 + 0 x 2 7 x + 1 x 4 + 1 2 x 2 1 2 x 2 7 x + 1 1 2 x 2 1 4 7 x + 5 4 Then f ( x ) = g ( x )( 1 2 x 2 1 4 ) + ( 7 x + 5 4 ) . (c) f ( x ) = 2 x 4 + x 2 x + 1 and g ( x ) = 2 x 1 in Z 5 [ x ]. 2 x 1 ¢ x 3 + 3 x 2 + 2 x + 3 2 x 4 + x 2 x + 1 2 x 4 x 3 x 3 + x 2 x + 1 x 3 3 x 2 4 x 2 x + 1 4 x 2 2 x x + 1 x 3 4 In the second step, when we divide the leading term of the divisor, 2 x , into x 3 , we obtain (1 · 2 1 ) x 2 . Since 2 · 3 = 1 in Z 5 , then 2 1 = 3 and we obtain 3 x 2 for the term in the quotient. Similarly, in the fourth step, we obtain 1 · 2 1 = 3. The result of the division is f ( x ) = g ( x )( x 3 + 3 x 2 + 2 x + 3) + 4 . 42

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This note was uploaded on 10/20/2009 for the course MATH 302 taught by Professor Edwards during the Spring '03 term at CSU Fullerton.

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4.1%20Polynomial%20Arithmetic%20II - Math 302 Modern...

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