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Unformatted text preview: Math 135, Winter 2010 Assignment 10 Solutions [4pts] (1) Let f ( x ) = 5 x 3 + 2 x 2 x 1 and g ( x ) = 2 x 2 + 2 x 2 be polynomials in Q [ x ]. Compute: (A) f ( x ) + g ( x ) Solution: f ( x ) = 5 x 3 + 2 x 2 x 1 g ( x ) = 2 x 2 + 2 x 2 f ( x ) + g ( x ) = 5 x 3 + 4 x 2 + x 3 So f ( x ) + g ( x ) = 5 x 3 + 4 x 2 x 3. (B) f ( x ) g ( x ) Solution: f ( x ) = 5 x 3 + 2 x 2 x 1 g ( x ) = 2 x 2 + 2 x 2 f ( x ) g ( x ) = 5 x 3 + 3 x + 1 So f ( x ) g ( x ) = 5 x 3 3 x + 1 (C) f ( x ) g ( x ) Solution: f ( x ) g ( x ) = (5 x 3 + 2 x 2 x 1)(2 x 2 + 2 x 2) = 5 x 3 (2 x 2 + 2 x 2) + 2 x 2 (2 x 2 + 2 x 2) x (2 x 2 + 2 x 2) (2 x 2 + 2 x 2) = 10 x 5 + 10 x 4 10 x 3 + 4 x 4 + 4 x 3 4 x 2 2 x 3 2 x 2 + 2 x 2 x 2 2 x + 2 = 10 x 5 + 14 x 4 8 x 3 8 x 2 + 2 So f ( x ) g ( x ) = 10 x 5 + 14 x 4 8 x 3 8 x 2 + 2. (D) the quotient and remainder when f ( x ) is divided by g ( x ). Solution: 5 2 x 3 2 2 x 2 + 2 x 2 5 x 3 + 2 x 2 x 1 5 x 3 + 5 x 2 5 x 3 x 2 + 4 x 1 3 x 2 3 x + 3 7 x 4 So the quotient is 5 2 x 3 2 and the remainder is 7 x 4 1 [4pts] (2) Let f ( x ) = x 3 + 2 x 2 + x + 4 and g ( x ) = 3 x 2 + 3 x + 1 be polynomials in Z 5 [ x ]. Compute: (A) f ( x ) + g ( x ) Solution: f ( x ) = x 3 + 2 x 2 + x + 4 g ( x ) = 3 x 2 + 3 x + 1 f ( x ) + g ( x ) = x 3 + 5 x 2 + 4 x + 5 = x 3 + 4 x So f ( x ) + g ( x ) = x 3 + 4 x . (B) f ( x ) g ( x ) Solution: f ( x ) = x 3 + 2 x 2 + x + 4 g ( x ) = 3 x 2 + 3 x + 1 f ( x ) g ( x ) = x 3 + 4 x 2 + 3 x + 3 So f ( x ) g ( x ) = x 3 + 4 x 2 + 3 x + 3. (C) f ( x ) g ( x ) Solution: f ( x ) g ( x ) = ( x 3 + 2 x 2 + x + 4)(3 x 2 + 3 x + 1) = x 3 (3 x 2 + 3 x + 1) + 2 x 2 (3 x 2 + 3 x + 1) + x (3 x 2 + 3 x + 1) + 4(3 x 2 + 3 x + 1) = 3 x 5 + 3 x 4 + 3 x 3 + x 4 + x 3 + 2 x 2 + 3 x 3 + 3 x 2 + x + 2 x 2 + 2 x + 4 = 3 x 5 + 4 x 4 + 2 x 2 + 3 x + 4...
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This note was uploaded on 04/13/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.
 Spring '08
 ANDREWCHILDS
 Polynomials

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