Alg_Complete

# Edutermsaspx college algebra b x 2 25 5x x2 again

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Unformatted text preview: ! (a) x2 - 2 x - 8 x 2 - 9 x + 20 We’ll first factor things out as completely as possible. Remember that we can’t cancel anything at this point in time since every term has a “+” or a “-” on one side of it! We’ve got to factor first! x2 - 2 x - 8 ( x - 4 ) ( x + 2) = x 2 - 9 x + 20 ( x - 5)( x - 4 ) At this point we can see that we’ve got a common factor in both the numerator and the denominator and so we can cancel the x-4 from both. Doing this gives, x2 - 2 x - 8 x + 2 = x 2 - 9 x + 20 x - 5 This is also all the farther that we can go. Nothing else will cancel and so we have reduced this expression to lowest terms. [Return to Problems] © 2007 Paul Dawkins 42 http://tutorial.math.lamar.edu/terms.aspx College Algebra (b) x 2 - 25 5x - x2 Again, the first thing that we’ll do here is factor the numerator and denominator. x 2 - 25 ( x - 5 ) ( x + 5 ) = 5x - x2 x (5 - x ) At first glance it looks there is nothing that will cancel. Notice however that there is a term in the denominator that is almost the same as a term in the numerator except all the signs are the opposite. We can use the following...
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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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