So we can combine up the middle two regions and the

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Unformatted text preview: the points in that region would satisfy the inequality. This means that all we need to do is look up at the number line above. If the test point from a region satisfies the inequality then that region is part of the solution. If the test point doesn’t satisfy the inequality then that region isn’t part of the solution. Now, also notice that any value of x that will satisfy the original inequality will also satisfy the inequality from Step 2 and likewise, if an x satisfies the inequality from Step 2 then it will satisfy the original inequality. So, that means that all we need to do is determine the regions in which the polynomial from Step 2 is negative. For this problem that is only the middle region. The inequality and interval notation for the solution to this inequality are, © 2007 Paul Dawkins 130 http://tutorial.math.lamar.edu/terms.aspx College Algebra ( -2,5) -2 < x < 5 Notice that we do need to exclude the endpoints since we have a strict inequality (< in this case) in the inequality. Okay, that seems like a long proc...
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