There was lots of explanation in the previous example

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Unformatted text preview: to negative) then it MUST go through zero! So, that means that these two numbers ( x = 5 and x = -2 ) are the ONLY places where the polynomial can change sign. The number line is then divided into three regions. In each region if the inequality is satisfied by one point from that region then it is satisfied for ALL points in that region. If this wasn’t true (i.e it was positive at one point in the region and negative at another) then it must also be zero somewhere in that region, but that can’t happen as we’ve already determined all the places where the polynomial can be zero! Likewise, if the inequality isn’t © 2007 Paul Dawkins 129 http://tutorial.math.lamar.edu/terms.aspx College Algebra satisfied for some point in that region that it isn’t satisfied for ANY point in that region. This leads us into the next step. Step 4 : Graph the points where the polynomial is zero (i.e. the points from the previous step) on a number line and pick a test point from each of the regions. Plug each of these test points into the polynomial and determine...
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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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