MATH 121 - WRITTEN EXERCISES-MODULE 1

8 x x 3 2 x x 4x 3x 1 8x x 3

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Unformatted text preview: nomial (only two terms) 3 3 2 2 3 3 2 2 48.) - ( 8 x + x − 3) + ( 2 x + x ) − ( 4x + 3x − 1) = -8x − x + 3 + 2 x + x − 4x − 3x + 1 =-6x 3 − 3x 2 − 4x + 4 50.) (5m – 6) (3m + 4) = 5m(3m) + 5m(4) – 6(3m) – 6(4) = 15m 2 + (20m − 18m) − 24 = 15m 2 + 2m − 24 68.) � y − 1) + z �= (4y+z-1)(4y+z-1) (4 � � = (4y (4y ) + 4y ( z ) + 4y ( −1) + z (4y) + z ( z ) + z ( −1) − 1(4y) − 1( z) − 1( −1)) = (16 y 2 + 8 yz − 8 y + z 2 − 2 z + 1) = 16 y 2 − 8 y + 1 + 8 yz − 2 z + z 2 2 80.) ( 3 p + 5 ) = (3 p + 5)(3 p + 5) = (3 p (3 p ) + 3 p (5) + 5(3 p) + 5(5)) = 9 p 2 + 30 p + 25 2 3 3 4 2 86.) -z ( 9 − z ) + 4z ( 2 + 3z ) = -9 z − z + 8 z + 12 z = z 4 − 9 z 3 + 12 z 2 + 8 z -8r 3 s − 12r 2 s 2 + 20rs 3 -8r 3 s 12r 2 s 2 20rs 3 88. = − + 4rs 4rs 4rs 4rs 2 2 = -2r − 3rs + 5s Jimmy Morgan-MAT-121-OL-(1 Nov 09 thru 23 Jan 2010) R.4 Written Assignment 10. 28r 4 s 2 + 7 r 3 s − 35r 4 s 3 = 7 r 3 s ( 4rs + 1 − 5rs 2 ) 14.) ( 3 z + 2 ) ( z + 4 ) − ( z + 6 ) ( z + 4 ) = (3z(z)+3z(4)+2(z)+2...
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This note was uploaded on 04/07/2012 for the course MATH 121 taught by Professor Crockett during the Spring '12 term at Thomas Edison State.

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