Alg_Complete

# Return to problems b 3 x y 2 x 5 y dont

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Unformatted text preview: L binomials. Recall, ( 3x - 5) ( x + 2 ) = 3x ( x ) + 3 x ( 2 ) - 5 ( x ) - 5 ( 2 ) = 3x 2 + 6 x - 5 x - 10 = 3x 2 + x - 10 With radicals we multiply in exactly the same manner. The main difference is that on occasion we’ll need to do some simplification after doing the multiplication (a) ( x +2 )( x -5 ) ( x +2 )( ) x -5 = x ( x)-5 x + 2 x - 10 = x 2 - 3 x - 10 = x - 3 x - 10 © 2007 Paul Dawkins 21 http://tutorial.math.lamar.edu/terms.aspx College Algebra As noted above we did need to do a little simplification on the first term after doing the multiplication. [Return to Problems] ( (b) 3 x - y )(2 x -5 y ) Don’t get excited about the fact that there are two variables here. It works the same way! (3 x- y )(2 ) x - 5 y = 6 x 2 - 15 x y - 2 x y + 5 y 2 = 6 x - 15 xy - 2 xy + 5 y = 6 x - 17 xy + 5 y Again, notice that we combined up the terms with two radicals in them. [Return to Problems] ( )( (c) 5 x + 2 5 x - 2 ) Not much to do with this one. (5 )( ) x + 2 5 x - 2 = 25 x 2 - 10 x + 10 x - 4 =...
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## This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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