MATH135 - Assignment10 Solutions

MATH135 - Assignment10 Solutions - MATH 135 Fall 2007...

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Unformatted text preview: MATH 135 Fall 2007 Assignment #10 Hand-In Problems None Recommended Problems 1. Text, page 262, #5 2. Text, page 262, #8 3. Text, page 262, #22 4. Text, page 262, #24 5. Text, page 263, #30 6. Text, page 263, #31 7. Text, page 263, #32 8. Text, page 263, #37 9. Text, page 263, #41 10. Text, page 263, #48 11. Text, page 263, #56 12. Text, page 264, #58 13. Text, page 264, #62 14. Text, page 265, #99 15. Text, page 267, #119 16. Text, page 269, #158 MATH 135 Fall 2007 Assignment #10 Solutions Recommended Problems 1. Let f ( x ) = x 3 + 2 x + 2 and g ( x ) = x 4 + x 2 + x + 1. Then f ( x ) + g ( x ) = ( x 3 + 2 x + 2) + ( x 4 + x 2 + x + 1) = x 4 + x 3 + x 2 + 3 x + 3 = x 4 + x 3 + x 2 f ( x )- g ( x ) = ( x 3 + 2 x + 2)- ( x 4 + x 2 + x + 1) =- x 4 + x 3- x 2 + x + 1 = 2 x 4 + x 3 + 2 x 2 + x + 1 f ( x ) g ( x ) = ( x 3 + 2 x + 2)( x 4 + x 2 + x + 1) = x 7 + 2 x 5 + 2 x 4 + x 5 + 2 x 3 + 2 x 2 + x 4 + 2 x 2 + 2 x + x 3 + 2 x + 2 = x 7 + 3 x 5 + 3 x 4 + 3 x 3 + 4 x 2 + 4 x + 2 = x 7 + x 2 + x + 2 2. Let f ( x ) = x 5 + 2 x + 4 and g ( x ) = x 7 + 5 x 3 + 4. Then f ( x ) + g ( x ) = ( x 5 + 2 x + 4) + ( x 7 + 5 x 3 + 4) = x 7 + x 5 + 5 x 3 + 2 x + 8 = x 7 + x 5 + 5 x 3 + 2 x + 1 f ( x )- g ( x ) = ( x 5 + 2 x + 4)- ( x 7 + 5 x 3 + 4) =- x 7 + x 5- 5 x 3 + 2 x + 0 = 6 x 7 + x 5 + 2 x 3 + 2 x f ( x ) g ( x ) = ( x 5 + 2 x + 4)( x 7 + 5 x 3 + 4) = x 12 + 2 x 8 + 4 x 7 + 5 x 8 + 10 x 4 + 20 x 3 + 4 x 5 + 8 x + 16 = x 12 + 7 x 8 + 4 x 7 + 4 x 5 + 10 x 4 + 20 x 3 + 8 x + 16 = x 12 + 4 x 7 + 4 x 5 + 3 x 4 + 6 x 3 + x + 2 3. Using long division,- ix 2 + x + ( i- 2) ix 2 + x- 2 | x 4 + 0 x 3 + x 2 + x + 1 x 4- ix 3 + 2 ix 2 ix 3- 2 ix 2 + x ix 3 + x 2- 2 x (- 1- 2 i ) x 2 + 3 x + 1 (- 1- 2 i ) x 2 + ( i- 2) x + (4- 2 i ) (5- i ) x + (2 i- 3) Therefore, the quotient is- ix 2 + x + ( i- 2) and the remainder is (5- i ) x + (2 i- 3). 4. 2 x 2 + 4 x + 3 5 x 3 + 2 x 2 + 6 x + 3 | 3 x 5 + 3 x 4 + 0 x 3 + 4 x 2 + 4 x + 3 3 x 5 + 4 x 4 + 5 x 3 + 6 x 2 6 x 4 + 2 x 3 + 5 x 2 + 4 x 6 x 4 + x 3 + 3 x 2 + 5 x x 3 + 2 x 2 + 6 x + 3 x 3 + 6 x 2 + 4 x + 2 3 x 2 + 2 x + 1 Therefore, the quotient is 2 x 2 + 4 x + 3 and the remainder is 3 x 2 + 2 x + 1....
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This note was uploaded on 09/21/2008 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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MATH135 - Assignment10 Solutions - MATH 135 Fall 2007...

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