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Unformatted text preview: lude the endpoints in our solution since at this points we get zero for the inequality and
0 ³ 0 is a true inequality.
Here is the solution in both inequality and interval notation form. ¥ < x £ 2 3£ x < ¥ ( ¥, 2] © 2007 Paul Dawkins and
and [3, ¥ ) 131 http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 3 Solve x 4 + 4 x 3  12 x 2 £ 0 .
Solution
Again, we’ll just jump right into the problem. We’ve already got zero on one side so we can go
straight to factoring. x 4 + 4 x3  12 x 2 £ 0
x 2 ( x 2 + 4 x  12 ) £ 0
x 2 ( x + 6 )( x  2 ) £ 0
So, this polynomial is zero at x = 6 , x = 0 and x = 2 . Here is the number line for this
problem. First, notice that unlike the first two examples these regions do NOT alternate between positive
and negative. This is a common mistake that students make. You really do need to plug in test
points from each region. Don’t ever just plug in for the first region and then assume that the other
regions will alternate from that point.
Now, for our s...
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 Spring '12
 MrVinh

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