M17 LE1 Sem1 09-10 Solns

# M17 LE1 Sem1 09-10 Solns - Math 17 First Long Exam...

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Math 17 First Long Exam Solutions V7, W3, Y3 I. 1. FALSE 2. TRUE 3. FALSE 4. TRUE 5. FALSE II. 1. ( - 2 i 2009 - i 2011 ) - 1 , 000 , 001 = ( - 2 i - ( - i )) - 1 , 000 , 001 = ( - i ) - 1 , 000 , 001 = ( - 1) - 1 , 000 , 001 ( i ) - 1 , 000 , 001 = - 1 ( i ) 1 , 000 , 001 = - 1 i · - i - i = i 2. We are given the following: A and B are subsets of the universal set U ; n ( U ) = 25, n ( A ) = 22, n ( B ) = 21, and n ( A B ) = 23. n ( A B ) = n ( A ) + n ( B ) - n ( A B ) n ( A B ) = n ( A )+ n ( B ) - n ( A B ) = 22+21 - 23 = 20 n (( A B ) 0 ) = n ( U ) - n ( A B ) = 25 - 20 = 5 ( (( A B ) 0 )) = 2 n (( A B ) 0 ) = 2 5 = 32 III. (2 x 2 - 4 8) 4 = (2 x 4 ) 4 - 4(2 x 4 ) 3 ( 4 8) + 6(2 x 4 ) 2 ( 4 8) 2 - 4(2 x 4 )( 4 8) 3 + ( 4 8) 4 = 16 x 16 - 4(8 x 12 )( 4 8) + 6(4 x 8 )(2 2) - 4(2 x 4 )(4 4 2) + 8 = 16 x 16 - 32 x 12 4 8 + 48 x 8 2 - 32 x 4 4 2 + 8 IV. Quotient: - 1 4 x 2 + 3 x - 1 Remainder: - 9 x + 9 V. 1. - 27 x 7 y 3 + 93 x 5 y 5 - 75 x 3 y 7 = - 3 x 3 y 3 (9 x 4 - 31 x 2 y 2 + 25 y 4 ) = - 3 x 3 y 3 [(9 x 4 - 30 x 2 y 2 + 25 y 4 ) - x 2 y 2 ] = - 3 x 3 y 3 [(3 x 2 - 5 y 2 ) 2 - x 2 y 2 ] = - 3 x 3 y 3 [(3 x 2 - 5 y 2 - xy )(3 x 2 - 5 y 2 + xy )] = - 3 x 3 y 3 (3 x 2 - xy - 5 y 2 )(3 x 2 + xy - 5 y 2 ) 2. ( p 6 - 1) + (3 p 5 - 3 p 2 ) + (3 p 4 - 3 p ) = ( p 3 + 1)( p 3 - 1) + 3 p 2 ( p 3 - 1) + 3 p ( p 3 - 1) = ( p 3 - 1)[( p 3 + 1) + 3 p 2 + 3 p ] = ( p 3 - 1)( p 3 + 3 p 2 + 3 p + 1) = ( p - 1)( p 2 + p + 1)( p + 1) 3 VI.

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