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Unformatted text preview: ess, however, it really isn’t. There was lots of explanation in the
previous example. The remaining examples won’t be as long because we won’t need quite as
much explanation in them. Example 2 Solve x 2 - 5 x ³ -6 .
Okay, this time we’ll just go through the process without all the explanations and steps. The first
thing to do is get a zero on one side and factor the polynomial if possible. x2 - 5x + 6 ³ 0 ( x - 3)( x - 2 ) ³ 0
So, the polynomial will be zero at x = 2 and x = 3 . Notice as well that unlike the previous
example, these will be solutions to the inequality since we’ve got a “greater than or equal to” in
Here is the number line for this example. Notice that in this case we were forced to choose a decimal for one of the test points.
Now, we want regions were the polynomial will be positive. So, the first and last regions will be
part of the solution. Also, in this case, we’ve got an “or equal to” in the inequality and so we’ll
need to inc...
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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.
- Spring '12