Edutermsaspx college algebra b x 2 25 5x x2 again

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online