document.write( "Question 1090441: Simplify the inequality, identify any critical points, and graph its solution
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Algebra.Com's Answer #704889 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "We factor the denominator\r\n" );
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document.write( "\"x%5E3-2x%5E2%2Bx-2\"\r\n" );
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document.write( "Factor out x2 from the first two terms and\r\n" );
document.write( "factor out +1 from the last two terms:\r\n" );
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document.write( "\"x%5E2%28x%5E%22%22-2%29%2B1%28x-2%29\" \r\n" );
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document.write( "Factor out (x-2)\r\n" );
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document.write( " \"%28x%5E%22%22-2%29%28x%5E2%2B1%29\"\r\n" );
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document.write( "Write 1 as \"1%2F1\"\r\n" );
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document.write( "The LCD is (x-2)(x²+1)\r\n" );
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document.write( "Multiply each numerator and denominator by whatever\r\n" );
document.write( "factor(s) that are needed so the resulting denominator\r\n" );
document.write( "will become the LCD.\r\n" );
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document.write( "Combine all the numerators over the LCD:\r\n" );
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document.write( "Combining terms:\r\n" );
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document.write( "\"%28x%5E3%2Bx-2%29%2F%28%28x%5E%22%22-2%29%28x%5E2%2B1%29%29+%3E0\"\r\n" );
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document.write( "That numerator x3+x-2 obviously has zero 1.\r\n" );
document.write( "So we factor it using synthetic division:\r\n" );
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document.write( "1 | 1  0  1 -2\r\n" );
document.write( "  |    1  1  2\r\n" );
document.write( "    1  1  2  0\r\n" );
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document.write( "So we see that it factors as (x-1)(x2+x+2)\r\n" );
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document.write( "\"%28%28x%5E%22%22-1%29%28x%5E2%2Bx%2B2%29%29%2F%28%28x%5E%22%22-2%29%28x%5E2%2B1%29%29+%3E0\"\r\n" );
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document.write( "The critical numbers are real zeros of the numerator and\r\n" );
document.write( "zeros of the denominator.\r\n" );
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document.write( "The only real zero of the numerator is 1\r\n" );
document.write( "The only real zero of the denominator is 2\r\n" );
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document.write( "We place those critical numbers on a number line\r\n" );
document.write( "They cannot be solutions themselves because the \r\n" );
document.write( "inequality does not permit equality:\r\n" );
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document.write( "-----------o----o----------\r\n" );
document.write( "-1    0    1    2    3    4\r\n" );
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document.write( "We choose a test value in the interval left of 1\r\n" );
document.write( "The easiest one is 0 and substitute it into the\r\n" );
document.write( "inequality:\r\n" );
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document.write( "\"%28%280%5E%22%22-1%29%280%5E2%2B0%2B2%29%29%2F%28%280%5E%22%22-2%29%280%5E2%2B1%29%29+%3E0\"\r\n" );
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document.write( "\"%28%28-1%29%282%29%29%2F%28%28-2%29%281%29%29%3E0\"\r\n" );
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document.write( "\"1%3E0\"\r\n" );
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document.write( "That is true, so we shade the number line left of 1\r\n" );
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document.write( "<==========o----o----------\r\n" );
document.write( "-1    0    1    2    3    4\r\n" );
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document.write( "We choose a test value between 1 and 2. The easiest \r\n" );
document.write( "one is 1.5 and substitute it into the\r\n" );
document.write( "inequality:\r\n" );
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document.write( "\"%28%280.5%29%285.75%29%29%2F%28%28-0.5%29%283.25%29%29%3E0\"\r\n" );
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document.write( "\"%22-1.769...%22%3E0\"\r\n" );
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document.write( "That is false, so we do not shade the number line between\r\n" );
document.write( "1 and 2.  So we still have this graph:\r\n" );
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document.write( "<==========o----o----------\r\n" );
document.write( "-1    0    1    2    3    4\r\n" );
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document.write( "We choose a test value in the interval right of 2\r\n" );
document.write( "The easiest one is 3 and substitute it into the\r\n" );
document.write( "inequality:\r\n" );
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document.write( "\"%28%283%5E%22%22-1%29%283%5E2%2B3%2B2%29%29%2F%28%283%5E%22%22-2%29%283%5E2%2B1%29%29+%3E0\"\r\n" );
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document.write( "\"%28%282%29%2814%29%29%2F%28%281%29%2810%29%29%3E0\"\r\n" );
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document.write( "\"2.8%3E0\"\r\n" );
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document.write( "That is true, so we shade the number line right of 2\r\n" );
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document.write( "<==========o----o==========>\r\n" );
document.write( "-1    0    1    2    3    4\r\n" );
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document.write( "That is the graph of the solution.  In interval notation\r\n" );
document.write( "that is written:\r\n" );
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document.write( "Edwin
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